THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


GIFT  OF 
John  S.  Prell 


Works  of  Professor  Mansfield  Merriman. 


Published  by  JOHN  WILEY  &  SONS,  53  E.  Tenth 
Street,  New  York. 


A  TREATISE  ON  HYDRAULICS. 

Designed  as  a  Text- Book  for  Technical  Schools  and  for  the 
use  of  Engineers.  By  Professor  Mansfield  Merriman,  Lehigh 
University.  Fourth  edition,  revised 8vo,  cloth,  $3  50 

"  As  a  whole  this  book  is  the  most  valuable  addition  to  the  literature  of 
hydraulic  science  which  has  yet  appeared  in  America,  and  we  do  not  know 
of  any  of  equal  value  anywhere  else."—  Baitroad  Gazette. 

"With  a  tolerably  complete  knowledge  of  what  has  been  written  on 
Hydraulics  in  England,  France,  Germany,  United  States,  and  to  some  extent 
Italy,  I  have  no  hesitation  in  saying  that  I  hold  this  book  to  be  the  best 
treatise  for  students,  young  or  old,  yet  written.  It  better  presents  the 
primary  essentials  of  the  art."— From  CLEMENS  HEBSCHEL,  Hydraulic 
Engineer  of  the  Holyoke  Water  Power  Company. 

A  TEXT-BOOK  ON  THE  METHOD  OP  LEAST  SQUARES. 
By  Mansfield  Merriman,  C.E.,  Ph.D.,  Professor  of  Civil 
Engineering  in  Lehigh  University.  Fifth  revised  edition. 

8vo,  cloth,  2  00 

This  work  treats  of  the  law  of  probability  of  error,  the  ad- 
justment and  discussion  of  observations  arising  in  surveying, 
geodesy,  astronomy  and  physics,  and  the  methods  of  compar- 
ing their  degrees  of  precision.  Its  rules  and  tables  will  assist 
all  who  wish  to  make  accurate  measurements. 
"  This  is  a  very  useful  and  much  needed  text-book." — Science.  • 

"  Even  the  casoal  reader  cannot  fail  to  be  struck  with  the  value  which 
such  a  book  must  possess  to  the  working  engineer.  It  abounds  in  illustra- 
tions and  problems  drawn  directly  from  surveying,  geodesy  and  engineer- 
ing."— Engineering  Ntw. 

THE   MECHANICS  OF  MATERIALS  AND  OP  BEAMS, 
COLUMNS,    AND    SHAFTS. 

By  Professor  Mansfield  Merriman,  Lehigh  University,  South 

Bethlehem,  Pa. 

Fourth  edition  revised  and  enlarged.     8vo,  cloth,  interleaved,  3  50 

"We  cannot  commend  the  book  too  highly  to  the  consideration  of  all 
Professors  of  Applied  Mechanics  and  Engineering  and  Technical  Schools 
and  Colleges,  and  we  think  a  general  introduction  of  the  work  will  mark  an 
advance  in  the  rational  of  technical  instruction."— American  Engineer. 

"The  mathematical  deductions  of  the  laws  of  strength  and  stiffness  of 
beams,  supported,  fixed,  and  continuous,  under  compression,  tension  and 
torsion,  and  of  columns,  are  elegant  and  complete.  As  in  previous  books 
by  the  same  author,  plenty  of  practical,  original  and-  modera  examples  are 
introduced  as  problems.  "-Proceeding  Engineers"  Club  of  Philadelphia. 


A  TEXT-BOOK  ON  ROOFS  AND  BRIDGES. 

Being  the  course  of  instruction  given  by  the  author  to  the 
students  of  civil  engineering  in  Lehigh  University. 
To  be  completed  in  four  parts. 

PART  I.     STRESSES    IN    SIMPLE    TRUSSES.     By  Professor 
Mansfield  Merriman.     Third  edition.     8vo,  cloth $2  60 

"  The  author  gives  the  most  modern  practice  in  determining  the  stresses 
due  to  moving  loads,  taking  actual  typical  locomotive  wheel  loads,  and 
reproduces  the  Phoenix  Bridge  Co's  diagram  for  tabulating  wheel  move- 
ments. The  whole  treatment  is  concise  and  very  clear  and  elegant."— Bail- 
road  Gazette. 

PART  II.     GRAPHIC  STATICS.     By  Professors  Mansfield  Mer- 
riman and  Henry  S.  Jacoby.     Second  edition.      8vo,   cloth,  2  50 

"  The  plan  of  this  book  is  simple  and  easily  understood ;  and  as  the  treat- 
ment of  all  problems  is  graphical,  mathematics  can  scarcely  be  said  to  enter 
into  its  composition,  .fudging  from  our  own  correspondence,  it  is  a  work 
for  which  there  is  a  decided  demand  outside  of  technical  schools." 

—Engineering  News. 

PART  III.    BRIDGE  DESIGN.    In  Preparation. 

This  volume  is  intended  to  include  the  design  of  plate  girders, 
lattice  trusses,  and  pin-connected  bridges,  together  with  the 
proportioning  of  details,  the  whole  being  in  accordance  with 
the  best  modern  practice  and  especially  adapted  to  the  needs 
of  students. 

THE    FIGURE     OF     THE     EARTH.      An    Introduction    to 

Geodesy. 

By  Mansfield  Merriman,  Ph.D.,  formerly  Acting  Assistant 
United  States  Coast  and  Geodetic  Survey.  12mo,  cloth 1  50 

"  It  is  BO  far  popularized,  that  there  are  few  persons  of  ordinary  intelli- 
gence who  may  not  read  it  with  profit  and  certainly  great  interest.' '—Engi- 
neering News. 

"  A  clear  and  concise  introduction  to  the  science  of  geodesy.  The  book 
is  interesting  and  deals  with  the  subject  in  a  useful,  and  to  some  extent, 
popular  manner."— London  Engineering. 

A  TEXT-BOOK  ON  RETAINING  WALLS  AND  MASONRY 
DAMS. 

By  Professor  Mansfield  Merriman,  Lehigh  University. 

8vo,  cloth,  2  00 

This  work  is  designed  not  only  as  a  text-book  for  students 
but  also  for  the  use  of  civil  engineers.  It  clearly  sets  forth 
the  methods  of  computing  the  thrust  of  earth  against  walls, 
and  the  investigation  and  design  of  walls  and  dams  in  the 
most  economic  manner.  The  principles  and  formulas  are 
illustrated  by  numerous  numerical  examples. 


XEXT=BOOK: 


MECHANICS  OF  MATERIALS 


BEAMS,  COLUMNS,  AND   SHAFTS. 


BY 

MANSFIELD   MERRIMAN, 

PROFESSOR    OF    CIVIL   ENGINEERING   IN   LEHIGH   UNIVERSITY. 


FOURTH   EDITION, 


NEW    YORK: 
JOHN     WILEY    &    SONS, 

53  EAST  TENTH  STREET. 
1892. 

JOHJY  S.  PRELL 

Civil  &  Mechanical  Engineer. 

SAN  FRANCISCO,  CAL. 


COPYRIGHT,  1890, 
BY  MANSFIELD   MERRIMAN. 


DHtTjniojro  &  NBU,  FERRIS  BROS., 

Electrotype™,  Printers 

1  to  7  Hague  Street.  «,  Pearl  Street. 

New  York. 


New  York. 


•350 


PREFACE. 


The  following  pages  contain  an  elementary  course  of  study 
in  the  resistance  of  materials  and  the  mechanics  of  beams, 
columns  and  shafts,  designed  for  the  use  of  classes  in  technical 
schools  and  colleges.  It  should  be  preceded  by  a  good  train- 
ing in  mathematics  and  theoretical  mechanics,  and  be  followed 
by  a  special  study  of  the  properties  of  different  qualities  of 
materials,  and  by  detailed  exercises  in  construction  and  design. 

As  the  plan  of  the  book  is  to  deal  mainly  with  the  mechanics 
of  the  subject,  extended  tables  of  the  results  of  tests  on  different 
kinds  and  qualities  of  materials  are  not  given.  The  attempt, 
however,  has  been  made  to  state  average  values  of  the  quanti- 
ties which  express  the  strength  and  elasticity  of  what  may  be 
called  the  six  principal  materials.  On  account  of  the  great 
variation  of  these  values  in  different  grades  of  the  same  material 
the  wisdom  of  this  attempt  may  perhaps  be  questioned,  but 
the  experience  of  the  author  in  teaching  the  subject  during  the 
past  eleven  years  has  indicated  that  the  best  results  are  attained 
by  forming  at  first  a  definite  nucleus  in  the  mind  of  the  student, 
around  which  may  be  later  grouped  the  multitude  of  facts 
necessary  in  his  own  particular  department  of  study  and  work. 

As  the  aim  of  all  education  should  be  to  develop  the  powers 
of  the  mind  rather  than  impart  mere  information,  the  author 
has  endeavored  not  only  to  logically  set  forth  the  principles 
and  theory  of  the  subject,  but  to  so  arrange  the  matter  that 
students  will  be  encouraged  and  required  to  think  for  them- 
selves. The  problems  which  follow  each  article  will  be  found 


713471 


iv  PREFACE. 

useful  for  this  purpose.  Without  the  solution  of  many  numer- 
ical exercises  it  is  indeed  scarcely  possible  to  become  well 
grounded  in  theory. 

In  the  chapters  on  flexure  many  problems  relating  to  I  beams 
and  other  wrought  iron  shapes  are  presented.  The  subject  of 
continuous  beams  is  not  developed  to  its  full  extent,  but  it  is 
thought  that  enough  is  given  for  an  elementary  course.  The 
resistance  of  columns  has  been  treated  with  as  much  fullness  as 
now  appears  practicable  from  a  theoretical  point  of  view.  Con- 
siderable attention  has  been  paid  to  combined  stresses^and 
particularly  to  the  combination  of  torsion  and  flexure  in  shafts. 
A  new  formula  for  the  case  of  repeated  stresses  is  presented, 
and  the  discussions  regarding  the  effect  of  shocks  and  the  inter- 
nal work  in  beams  are  believed  to  be  novel.  The  attempt  has 
been  made  to  render  the  examples,  exercises,  and  problems  of 
a  practical  nature,  and  also  of  a  character  to  clearly  illustrate 
the  principles  of  the  theory  and  the  methods  of  investigation. 

The  present  edition  is  the  result  of  a  careful  revision,  many 
alterations  and  additions  having  been  made  in  order  to  render 
the  book  more  efficient  for  class  use.  The  answers  to  problems 
contained  in  the  key  to  the  first  edition  are  here  given  in  the 
appendix.  All  the  cuts  have  been  redrawn  and  several  new 
ones  inserted.  Lastly,  on  account  of  the  universal  approbation 
which  has  been  expressed  concerning  the  author's  experiment 
in  issuing  his  '  Roofs  and  Bridges '  with  each  alternate  leaf 
blank,  the  same  plan  has  been  here  adopted.  On  these  blank 
pages  students  may  record  their  solutions  of  the  problems  in  a 
permanent  form  that  will  be  of  great  value  to  them  in  subse- 
quent practical  work. 

MANSFIELD  MERRIMAN. 

BETHLEHEM,  PA.,  December,  1889. 


CONTENTS. 


CHAPTER  I. 
THE   RESISTANCE  AND   ELASTICITY  OF   MATERIALS. 

PAGE 

ART.    i.  AVERAGE  WEIGHTS i 

2.  STRESSES  AND  STRAINS, 3 

3.  EXPERIMENTAL  LAWS -t  .     .  6 

4.  ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY,       .  f 

5.  TENSION, 9 

6.  COMPRESSION, -   .        .13 

7.  SHEAR 15 

8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES,     .       .  17 

CHAPTER   II. 

PIPES,  CYLINDERS,  AND   RIVETED  JOINTS. 

ART.    9.    WATER  AND  STEAM  PIPES, 22 

10.    CYLINDERS  AND  SPHERES, 24 

n.    THICK  CYLINDERS, '      .       .       .26 

12.  INVESTIGATION  OF  RIVETED  JOINTS,        .       .       .       .28 

13.  DESIGN  OF  RIVETED  JOINTS 32 

14.  MISCELLANEOUS  EXERCISES, 34 

CHAPTER  III. 
CANTILEVER  BEAMS  AND  SIMPLE  BEAMS. 

ART.  15.  DEFINITIONS 36 

16.  REACTIONS  OF  THE  SUPPORTS,         .        .       .       .       .37 

17.  THE  VERTICAL  SHEAR 39 

18.  THE  BENDING  MOMENT 42 


Vi  CONTENTS. 

PAGE 

ART.  19.  INTERNAL  STRESSES  AND  EXTERNAL  FORCES,       .       .    45 

20.  EXPERIMENTAL  AND  THEORETICAL  LAWS,     .      " .       .48 

21.  THE  Two  FUNDAMENTAL  FORMULAS,     .       ,       .       .    49 

22.  CENTER  OF  GRAVITY  OF  CROSS-SECTIONS,     ...       .52 

23.  MOMENT  OF  INERTIA  OF  CROSS-SECTIONS,     .       .    .    .    53 

24.  THE  MAXIMUM  BENDING  MOMENT,         .       ....    54 

25.  THE  INVESTIGATION  OF  BEAMS,       .      V      .     %       .56 

26.  SAFE  LOADS  FOR  BEAMS,  .       .       .       .       .       .  •     .    58 

27.  DESIGNING  OF  BEAMS 59 

28.  THE  MODULUS  OF  RUPTURE, 61 

29.  COMPARATIVE  STRENGTHS ".62 

30.  WROUGHT  IRON  I  BEAMS 64 

31.  WROUGHT  IRON  DECK  BEAMS, 67 

32.  CAST  IRON  BEAMS 68 

33.  GENERAL  EQUATION  OF  THE  ELASTIC  CURVE,      .        .    70 

34.  DEFLECTION  OF  CANTILEVER  BEAMS,      .        .       .       .72 

35.  DEFLECTION  OF  SIMPLE  BEAMS, 74 

36.  COMPARATIVE  DEFLECTION  AND  STIFFNESS,  .       .        .77 

37.  RELATION  BETWEEN  DEFLECTION  AND  STRESS,    .        .    79 

38.  CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH,        .       .    80 

39.  SIMPLE  BEAMS  OF  UNIFORM  STRENGTH,        ...    83 


CHAPTER  IV. 

RESTRAINED   BEAMS  AND  CONTINUOUS  BEAMS. 

ART.  40.    BEAMS  OVERHANGING  ONE  SUPPORT 85 

41.  BEAMS    FIXED   AT    ONE    END   AND    SUPPORTED    AT 

THE  OTHER,        .        .        .        .-      .        .       .        .88 

42.  BEAMS  OVERHANGING  BOTH  SUPPORTS,         .       .        .91 

43.  BEAMS  FIXED  AT  BOTH  ENDS, 92 

44.  COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS,      .    94 

45.  GENERAL  PRINCIPLES  OF  CONTINUITY,   ....    96 

46.  PROPERTIES  OF  CONTINUOUS  BEAMS,     *       .  .99 

47.  THE  THEOREM  OF  THREE  MOMENTS,      *       *•       .       .102 

48.  CONTINUOUS  BEAMS  WITH  EQUAL  SPANS,     .       .        .103 

49.  CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS,        .        .  106 
£o.  REMARKS  ON  THE  THEORY  OF  FLEXURE,      .        .        .107 


CONTENTS. 

vii 

CHAPTER  V. 

THE  COMPRESSION  OF  COLUMNS, 

PACK 

ART.  51. 

CROSS-SECTIONS  OF  COLUMNS,  

.    Ill 

52. 

GENERAL  PRINCIPLES  

•  "3 

53- 

EULER'S  FORMULAS,    .... 

.  114 

54- 

HODGKINSON'S  FORMULAS,        . 

.  117 

55- 

GORDON'S  FORMULA  

.  119 

56. 

RANKINE'S  FORMULA,         

.    122 

57- 

RADIUS  OF  GYRATION  OF  CROSS-SECTIONS,    . 

.    124 

58. 

INVESTIGATION  OF  COLUMNS  

.    125 

59- 

SAFE  LOADS  FOR  COLUMNS,      

126 

60. 

DESIGNING  OF  COLUMNS  

.    127 

61. 

EXPERIMENTAL  RESULTS  

.    I29 

62. 

REMARKS  ON  THE  THEORY  OF  COLUMNS,      . 

•    131 

CHAPTER   VI. 

TORSION,   AND  SHAFTS   FOR  TRANSMITTING  POWER. 

ART.  63. 

THE  PHENOMENA  OF  TORSION  

•    135 

64. 

THE  FUNDAMENTAL  FORMULA  FOR  TORSION, 

•    136 

65. 

POLAR  MOMENTS  OF  INERTIA,          .... 

.   138 

66. 

THE  CONSTANTS  OF  TORSION  

•    139 

67. 

SHAFTS  FOR  TRANSMISSION  OF  POWER,  . 

.    140 

68. 

ROUND  SHAFTS,  .       .       .       ... 

.   141 

69. 

SQUARE  SHAFTS,         .               .       .       . 

.   142 

70. 

MISCELLANEOUS  EXERCISES,     .       .       .       .       . 

•   143 

CHAPTER  VII. 

COMBINED  STRESSES. 

ART.  71. 

CASES  OF  COMBINED  STRESSES,        .... 

.   144 

72. 

STRESSES  DUE  TO  TEMPERATURE  

.   145 

73- 

COMBINED  TENSION  AND  FLEXURE, 

.    I46 

74- 

COMBINED  COMPRESSION  AND  FLEXURE, 

.    I48 

75- 

SHEAR  COMBINED  WITH  TENSION  OR  COMPRESSION, 

.    ISO 

76. 

COMBINED  FLEXURE  AND  TORSION, 

.    152 

77- 

COMBINED  COMPRESSION  AND  TORSION,  . 

•    154 

viii  CONTENTS. 

PAGE 

ART.  78.    HORIZONTAL  SHEAR  IN  BEAMS, 155 

79.    MAXIMUM  INTERNAL  STRESSES  IN  BEAMS,     .       .       .158 


CHAPTER  VIII. 

APPENDIX  AND  TABLES. 

ART.  80.  SUDDEN  LOADS  AND  SHOCKS, 162 

81.  THE  RESILIENCE  OF  MATERIALS, 164 

82.  THE  FATIGUE  OF  METALS 166 

83.  WORKING  STRENGTHS  FOR  REPEATED  STRESSES,         .  167 

84.  THE  INTERNAL  WORK  IN  BEAMS, 171 

85.  ANSWERS  TO  PROBLEMS 174 

86.  TABLES  OF  CONSTANTS, 176 


MECHANICS  OF  MATERIALS. 


CHAPTER   I. 
THE   RESISTANCE  AND   ELASTICITY   OF   MATERIALS. 

ARTICLE  i.    AVERAGE  WEIGHTS. 

The  principal  materials  used  in  engineering  constructions 
are  timber,  brick,  stone,  cast  iron,  wrought  iron,  and  steel. 
The  following  table  gives  their  average  unit-weights  and  aver- 
age specific  gravities. 


Material. 

Average  Weight. 

Average 
Specific 
Gravity. 

Pounds  per 
Cubic  Foot. 

Kilos  per 
Cubic  Meter. 

Timber 

40 

600 

0.6 

Brick 

125 

2000 

1      2.0 

Stone 

1  60 

2  560 

2.6 

Cast  Iron 

450 

7  200 

7-2 

Wrought  Iron 

480 

7700 

7-7 

Steel 

490 

7  800 

7.8 

These  weights,  being  mean  or  average  values,  should  be  care- 
fully memorized  by  the  student  as  a  basis  for  more  precise 
knowledge,  but  it  must  be  noted  that  they  are  subject  to  more 
or  less  variation  according  to  the  quality  of  the  material. 
Brick,  for  instance,  may  weigh  as  low  as  100,  or  as  high  as  150 
pounds  per  cubic  foot,  according  as  it  is  soft  or  hard  pressed. 


2  RESISTANCE   AND   ELASTICITY   OF   MATERIALS.      CH.  I. 

Unless  otherwise  stated  the  above  average  values  will  be  used 
in  the  examples  and  problems  of  this  book.  In  all  engineering 
reference  books  are  given  tables  showing  the  unit-weights  for 
different  qualities  of  the  above  six  principal  materials,  and  also 
for  copper,  lead,  glass,  cements,  and  other  materials  used  in 
construction. 

For  computing  the  weights  of  bars,  beams,  and  pieces  of  uni- 
form cross-section,  the  following  approximate  simple  rules  will 
often  be  found  convenient. 

A  wrought  iron  bar  one  square  inch  in  section  and  one 
yard  long  weighs  ten  pounds. 

Steel  is  about  two  per  cent  heavier  than  wrought  iron. 

Cast  iron  is  about  six  per  cent  lighter  than  wrought  iron. 

Stone  is  about  one-third  the  weight  of  wrought  iron. 

Brick  is  about  one-fourth  the  weight  of  wrought  iron. 

Timber  is  about  one-twelfth  the  weight  of  wrought  iron. 
For  example,  consider  a  bar  of  wrought  iron  if  X  3  inches  and 

12  feet  long;   its  cross-section  is  4.5  square  inches,  hence  its 
weight   is  45  X  4  =  1 80  pounds.     A  steel   bar  of   the   same 
dimensions  will  weigh   180  +  0.02  X  1 80  =  about  184  pounds, 
and  a  cast  iron  bar  will  weigh  180  —  0.06  X  180  =  about  169 
pounds. 

By  reversing  the  above  rules  the  cross-sections  of  bars  are 
readily  computed  from  their  weights  per  yard.  Thus,  if  a  stick 
of  timber  15  feet  long  weigh  120  pounds,  its  weight  per  yard  is 
24  pounds,  and  its  cross-section  is  12  X  2.4  =  about  28.8  square 
inches. 

Problem  i.  How  many  square  inches  in  the  cross-section  of  a 
wrought  iron  railroad  rail  weighing  24  pounds  per  linear  foot  ? 
In  a  steel  rail?  In  a  wooden  beam? 

Prob.  2.  Find  the  weights  of  a  wooden  beam  6x8  inches  in 
section  and  13  feet  long,  of  a  steel  bar  one  inch  in  diameter  and 

13  feet  long,  and  of  a  stone  block  18  X  24  inches  and  9  feet 
long. 


ART.  2.  STRESSES  AND  DEFORMATIONS.  3 

ART.  2.    STRESSES  AND  DEFORMATIONS. 

A  '  stress '  is  a  force  which  acts  in  the  interior  of  a  body  and 
resists  the  external  forces  which  tend  to  change  its  shape.  If 
a  weight  of  400  pounds  be  suspended  by  a  rope,  the  stress  in 
the  rope  is  400  pounds.  This  stress  is  accompanied  by  an 
elongation  of  the  rope,  which  increases  until  the  internal  molec- 
ular stresses  or  resistances  are  in  equilibrium  with  the  exterior 
weight.  Stresses  are  measured  in  pounds,  tons,  or  kilograms. 
A  '  unit-stress '  is  the  amount  of  stress  on  a  unit  of  area ;  this  is 
expressed  either  in  pounds  per  square  inch,  or  in  kilograms  per 
square  centimeter.  Thus,  if  a  rope  of  two  square  inches  cross- 
section  sustains  a  stress  of  400  pounds,  the  unit-stress  is  200 
pounds  per  square  inch,  for  the  total  stress  must  be  regarded 
as  distributed  over  the  two  square  inches  of  cross-section. 

A  'deformation  '  is  the  amount  of  change  of  shape  of  a  body 
caused  by  the  stress.  For  instance,  if  a  load  be  put  on  a 
column  its  length  is  shortened,  and  the  amount  of  shortening 
is  a  deformation.  So  in  the  case  of  the  rope,  the  amount  of 
elongation  is  a  deformation.  Deformations  are  generally  meas- 
ured in  inches,  or  centimeters. 

The  word  '  strain  '  is  often  used  in  technical  literature  as 
synonymous  with  stress,  and  sometimes  it  is  also  used  to  desig- 
nate the  deformation,  or  change  of  shape.  On  account  of  this 
ambiguity  the  word  will  not  be  employed  in  this  book. 

Three  kinds  of  simple  stress  are  produced  by  forces  which 
tend  to  change  the  shape  of  a  body.  They  are, 

Tensile,  tending  to  pull  apart,  as  in  a  rope. 
Compressive,  tending  to  push  together,  as  in  a  column. 
Shearing,  tending  to  cut  across,  as  in  punching  a  plate. 

The  nouns  corresponding  to  these  three  adjectives  are  Tension, 
Compression,  and  Shear.     The  stresses  which  occur  in  beams, 


4  RESISTANCE   AND   ELASTICITY   OF   MATERIALS.      CH.  I. 

columns,  and  shafts  are  of  a  complex  character,  but  they  may 
always  be  resolved  into  the  three  kinds  of  simple  stress.  The 
first  effect  of  a  stress  is  to  cause  a  deformation  in  the  body. 
This  deformation  receives  a  special  name  according  to  the  kind 
of  stress  which  produces  it.  Thus, 

Tension  produces  an  elongation. 
Compression  produces  a  shortening. 
Shear  produces  a  detrusion. 

This  change  of  shape  is  resisted  by  the  stresses  between  the 
molecules  of  the  body,  and  as  soon  as  these  internal  resistances 
balance  the  exterior  forces  the  change  of  shape  ceases  and  the 
body  is  in  equilibrium.  But  if  the  external  forces  be  increased 
far  enough  the  molecular  resistances  are  finally  overcome  and 
the  body  breaks  or  ruptures. 

In  any  case  of  simple  stress  in  a  body  in  equilibrium  the 
total  internal  stresses  or  resistances  must  equal  the  external 
applied  force.  Thus,  in  the  above  instance  of  a  rope  from 
which  a  weight  of  400  pounds  is  suspended,  let  it  be  imagined 
to  be  cut  at  any  section ;  then  equilibrium  can  only  be  main- 
tained by  applying  at  that  section  an  upward  force  of  400 
pounds ;  hence  the  stresses  in  that  section  must  also  equal  400 
pounds.  In  general,  if  a  steady  force  P  produce  either  ten- 
sion, compression,  or  shear,  the  total  stress  produced  is  also  P, 
for  if  not  equilibrium  does  not  obtain.  In  such  cases,  then,  the 
word  '  stress '  may  be  used  to  designate  the  external  force  as 
well  as  the  internal  resistances. 

Tension  and  Compression  are  similar  in  character  but  differ 
in  regard  to  direction.  A  tensile  stress  in  a  bar  occurs  when 
two  forces  of  equal  intensity  act  upon  its  ends,  each  in  a  direc- 
tion away  from  the  other.  In  compression  the  direction  of  the 
forces  is  reversed  and  each  acts  toward  the  bar.  Evidently  a 
simple  tensile  or  compressive  stress  in  a  bar  is  to  be  regarded 
as  evenly  distributed  over  the  area  of  its  cross-section,  so  that 


ART.  2.  STRESSES   AND   DEFORMATIONS.  5 

if  P  be  the  total  stress  in  pounds  and  A  the  area  of  the  cross- 

p 
section  in  inches,  the  unit-stress  is  -j  in  pounds  per  square  inch. 

Shear  requires  the  action  of  two  forces  exerted  in  parallel 
planes  and  very  near  together,  like  the  forces  in  a  pair  of 
shears,  from  which  analogy  the  name  is  derived.  Here  also 
the  total  shearing  stress  P  is  to  be  regarded  as  distributed 

p 
uniformly  over  the  area  A,  so  that  the  unit-stress  is    j-     And 

conversely  if  5  represent  the  uniform  unit-stress  the  total  stress 
Pis  AS. 

In  any  case  of  simple  stress  acting  on  a  body  let  P  be  the 
total  stress,  A  the  area  over  which  it  is  uniformly  distributed, 
and  5  the  unit-stress.  Then, 

(i)  P=AS. 

Also  let  A.  be  the  total  linear  deformation  produced  by  the 
stress,  /  the  length  of  the  bar,  and  s  the  deformation  per  unit 
of  length.  Then  this  deformation  is  to  be  regarded  as  uni- 
formly distributed  over  the  distance  /,  so  that  also, 

(i)'  A  =  Is. 

The  laws  implied  in  the  statement  of  these  two  formulas  are 
confirmed  by  experiment,  if  the  stress  be  not  too  great. 

Unit-stress  in  general  will  be  denoted  by  S,  whether  it  be 
tension,  compression,  or  shear.  St  will  denote  tensile  unit- 
stress,  Sc  compressive  unit-stress,  and  vSs  shearing  unit-stress, 
when  it  is  necessary  to  distinguish  between  them. 

Prob.  3.  A  wrought  iron  rod  i  J  inches  in  diameter  breaks 
under  a  tension  of  67  500  pounds.  Find  the  breaking  unit- 
stress. 

Prob.  4.  If  a  wooden  bar  i  x  3  inches  breaks  under  a  tensile 
stress  of  30  ooo  pounds,  what  stress  will  break  a  bar  2x3^ 
inches? 


RESISTANCE   AND   ELASTICITY   OF   MATERIALS.      CH.  I. 


ART.  3.    EXPERIMENTAL  LAWS. 

Numerous  tests  or  experiments  have  been  made  to  ascertain 
the  strength  of  materials  and  the  laws  that  govern  stresses  and 
deformations.  The  resistance  of  a  rope,  for  instance,  may  be 
investigated  by  suspending  it  from  one  end  and  applying 
weights  to  the  other.  As  the  weights  are  added  the  rope  will 
be  seen  to  stretch  or  elongate,  and  the  amotint  of  this  deforma- 
tion may  be  measured.  When  the  load  is  made  great  enough 
the  rope  will  break,  and  thus  its  ultimate  tensile  stress  is 
known.  For  stone,  iron,  or  steel,  special  machines,  known  as 
testing  machines,  have  been  constructed  by  which  the  effect  of 
different  stresses  on  different  qualities  and  forms  of  materials 
may  be  accurately  measured. 

All  experiments,  and  all  experience,  agree  in  establishing  the 
five  following  laws  for  cases  of  simple  tension  and  compression, 
which  may  be  regarded  as  the  fundamental  principles  of  the 
science  of  the  strength  of  materials. 

(A) — When  a  small  stress  is  applied  to  a  body  a  small  de- 
formation is  produced,  and  on  the  removal  of  the  stress 
the  body  springs  back  to  its  original  form.  For  small 
stresses,  then,  materials  may  be  regarded  as  perfectly 
elastic. 

(B) — Under  small  stresses  the  deformations  are  approxi- 
mately proportional  to  the  forces,  or  stresses,  which  pro- 
duce them,  and  also  approximately  proportional  to  the 
length  of  the  bar  or  body. 

(C] — When  the  stress  is  great  enough  a  deformation  is 
produced  which  is  partly  permanent,  that  is,  the  body 
does  not  spring  back  entirely  to  its  original  form  on  re- 
moval of  the  stress.  This  permanent  part  is  termed  a 
set.  In  such  cases  the  deformations  are  not  proportional 
to  the  stresses. 


ART.  4.    ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY.       7 

(D) — When  the  stress  is  greater  still  the  deformation 
rapidly  increases  and  the  body  finally  ruptures. 

(£) — A  sudden  stress,  or  shock,  is  more  injurious  than  a 
steady  stress  or  than  a  stress  gradually  applied. 

The  words  small  and  great,  used  in  stating  these  laws,  have,  as 
will  be  seen  later,  very  different  values  and  limits  for  different 
kinds  of  materials  and  stresses. 

The 'ultimate  strength '  of  a  material  under  tension,  com- 
pression, or  shear,  is  the  greatest  unit-stress  to  which  it  can  be 
subjected.  This  occurs  at  or  shortly  before  rupture,  and  its 
value  is  very  different  for  different  materials.  Thus  if  a  bar 
whose  cross-section  is  A  breaks  under  a  tensile  stress  P,  the 
ultimate  tensile  strength  of  the  material  is  P -i-  A. 

Prob.  5.  If  the  ultimate  strength  of  wrought  iron  is  55000 
pounds  per  square  inch,  what  tension  will  rupture  a  bar  6  feet 
long  which  weighs  60  pounds  ? 

Prob.  6.  If  a  bar  i  inch  in  diameter  and  8  feet  long  elon- 
gates 0.05  inch  under  a  stress  of  15000  pounds,  how  much, 
according  to  law  (£\  will  a  bar  of  the  same  size  and  material 
elongate  whose  length  is  12  feet  and  stress  30000  pounds? 

ART.  4.    ELASTIC  LIMIT  AND  COEFFICIENT  OF  ELASTICITY. 

The  '  elastic  limit '  is  that  unit-stress  at  which  the  permanent 
set  is  first  visible  and  within  which  the  stress  is  directly  propor- 
tional to  the  deformation.  For  stresses  less  than  the  elastic 
limit  bodies  are  perfectly  elastic,  resuming  their  original  form 
on  removal  of  the  stress.  Beyond  the  elastic  limit  a  permanent 
alteration  of  shape  occurs,  or,  in  other  words,  the  elasticity  of 
the  material  has  been  impaired.  It  is  a  fundamental  rule  in  all 
engineering  constructions  that  materials  can  not  safely  be 
strained  beyond  their  elastic  limit. 

The  '  coefficient  of  elasticity  '  of  a  bar  for  tension,  compres- 
sion, or  shearing,  is  the  ratio  of  the  unit-stress  to  the  unit- 


8  RESISTANCE  AND   ELASTICITY   OF   MATERIALS.      CH.  I. 

deformation,  provided  the  elastic  limit  of  the  material  be  not 
exceeded.  Let  5  be  the  unit-stress,  s  the  unit-deformation,  and 
E  the  coefficient  of  elasticity.  Then  by  the  definition, 

(2)  E  =  -          and         5  =  £s. 

By  law  (B)  the  quantity  E  is  a  constant  for  each  material,  until 
S  reaches  the  elastic  limit.  Beyond  this  limit  s  increases  more 
rapidly  than  5  and  the  ratio  is  no  longer  constant.  Equation 
(2)  is  a  fundamental  one  in  the  science  of  the  strength  of 
materials.  Since  E  varies  inversely  with  s,  the  coefficient  of 
elasticity  may  be  regarded  as  a  measure  of  the  stiffness  of  the 
material.  The  stiffer  the  material  the  less  is  the  change  in 
length  under  a  given  stress  and  the  greater  is  E.  The  values 
of  E  for  materials  have  been  determined  by  experiments  with 
testing  machines  and  their  average  values  will  be  given  in  the 
following  articles.  E  is  necessarily  expressed  in  the  same  unit 
as  the  unit-stress  S.  Some  authors  give  the  name  '  modulus  of 
elasticity '  to  the  quantity  E. 

Another  definition  of  the  coefficient  of  elasticity  for  the  case 
of  tension  is  that  it  is  the  unit-stress  which  would  elongate  a 
bar  to  double  its  original  length,  provided  that  this  could  be 
done  without  exceeding  the  elastic  limit.  That  this  defini- 
tion is  in  agreement  with  (2)  may  be  shown  by  regarding  a  bar 
of  length  /  which  elongates  the  amount  A  under  the  unit- 

r>  .    i 

stress  -r.     Here  the  unit-elongation  is  7  and  (2)  becomes, 

P       A       PI 

(2>  E  =  A*1  =  AZ 

p 

and  if  A  be  equal  to  /,  E  is  the  same  as  the  unit-stress  -r* 

A 

Prob.  7.  Find  the  coefficient  of  elasticity  of  a  bar  of  wrought 
iron  i£  inches  in  diameter  and  16  feet  long  which  elongates  -J 
inch  under  a  tensile  stress  of  20  ooo  pounds. 


ART.  5. 


TENSION. 


9 


Prob.  8.  If  the  coefficient  of  elasticity  of  cast  iron  is  1 5  ooo  ooo 
pounds  per  square  inch,  how  much  will  a  bar  2X3  inches  and 
6  feet  long  stretch  under  a  tension  of  5  ooo  pounds  ? 

ART.  5.    TENSION. 

The  phenomena  of  tension  observed  when  a  gradually  in- 
creasing stress  is  applied  to  a  bar,  are  briefly  as  follows  :  When 
the  unit-stress  5  is  less  than  the  elastic  limit  Se ,  the  unit-elon- 
gation s  is  small  and  proportional  to  5.  Within  this  limit  the 
ratio  of  S  to  s  is  the  coefficient  of  elasticity  of  the  material. 
After  passing  the  elastic  limit  the  bar  rapidly  elongates  and 
this  is  accompanied  by  a  reduction  in  area  of  its  cross-section. 
Finally  when  6"  reaches  the  ultimate  tensile  strength  St,  the 
bar  tears  apart.  Usually  St  is  the  maximum  unit-stress  on  the 
bar,  but  in  some  cases  the  unit-stress  reaches  a  maximum 
shortly  before  rupture  occurs. 

The  constants  of  tension  for  timber,  cast  iron,  wrought  iron 
and  steel  are  given  in  the  following  table.  The  values  are 
average  ones* and  are  liable  to  great  variations  for  different 
grades  and  qualities  of  materials.  Brick  and  stone  are  not 
here  mentioned,  as  they  are  rarely  or  never  used  in  tension. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Tensile 
Strength,  St  . 

Ultimate 
Elongation,  j. 

Pounds  per  square 
inch. 

Pounds   per 
square  inch. 

Pounds  per 
square    inch. 

Inches   per 
linear  inch. 

Timber, 

i  500  ooo 

3  ooo 

IO  OOO 

O.OI5 

Cast  Iron, 

15  ooo  ooo 

6000 

20000 

O.OO5 

Wrought  Iron, 

25  ooo  ooo 

25  ooo 

55000 

O.2O 

Steel, 

30  ooo  ooo 

50  ooo 

IOO  OOO 

O.  IO 

The  values  of  the  coefficients  of  elasticity,  elastic  limits,  and 
breaking  or  ultimate  strengths  are  given  in  pounds  per  square 
inch  of  the  original  cross-section  of  the  bar.  The  ultimate 
elongations  are  in  fractional  parts  of  the  original  length,  or  they 


10  RESISTANCE   AND   ELASTICITY   OF   MATERIALS.       CH.  I. 


are  the  elongations  per  linear  unit ;  these  should  be  regarded 
as  very  rough  averages,  since  they  are  subject  to  great  variations 
depending  on  the  shape,  size,  and  quality  of  the  specimen. 

The  ultimate  elongation,  together  with  the  reduction  in  area 
of  the  cross-section,  furnishes  the  means  of  judging  of  the  duc- 
tility of  the  material.  The  reduction  of  area  in  cast  iron  and  in 
many  varieties  of  steel  is  scarcely  perceptible,  while  in  other 
varieties  of  steel  and  in  wrought  iron  it  may  be  as  high  as  0.4 
of  the  original  section. 

A  graphical  illustration  of  the  principal  phenomena  of  tension 
is  given  in  Fig.  I.  The  unit-stresses  are  taken  as  ordinates  and 

100,000 
90,000 


the  unit-elongations  as  abscissas.  For  each  unit-stress  the  cor- 
responding unit-elongation  as  found  by  experiment  is  laid  off, 
and  curves  drawn  through  the  points  thus  determined.  The 
curve  for  each  of  the  materials  is  a  straight  line  from  the  origin 
until  the  elastic  limit  is  reached,  as  should  be  the  case  accord- 
ing to  the  law  (B}.  The  tangent  of  the  angle  which  this  line 
makes  with  the  axis  of  abscissas  is  equal  to  5  -f-  s,  which  is  the 
same  in  value  as  the  coefficient  of  elasticity  of  the  material. 
At  the  elastic  limit  a  sudden  change  in  the  curve  is  noticed  and 
the  elongation  rapidly  increases.  The  termination  of  the  curve 


ART.  5. 


TENSION. 


II 


indicates  the  point  of  rupture.  These  curves  show  more 
plainly  to  the  eye  than  the  values  in  the  table  can  do  the 
differences  in  the  properties  of  the  materials.  It  will  be  seen 
that  the  elastic  limit  is  not  a  well  defined  point,  but  that  its 
value  is  more  or  less  uncertain,  particularly  for  cast  iron  and 
timber.  It  should  be  also  clearly  understood  that  individual 
curves  for  special  cases  would  often  show  marked  variations 
from  their  mean  forms  as  represented  in  the  diagram. 

As  a  particular  example  a 
tensile  test  of  a  wrought  iron 
bar  f  inches  in  diameter  and 
12  inches  long  made  at  the 
Pencoyd  Iron  Works  will  be 
considered.  In  the  first  col- 
umn of  the  following  table 
are  given  the  total  stresses 
which  were  successively  ap- 
plied, in  the  second  the 
stresses  per  square  inch,  in 
the  third  the  total  elonga- 
tions, and  in  the  fourth  the 
elongations  or  sets  after  re- 
moval of  the  stress.  The 
unit-elongations  are  found  by 
dividing  those  in  the  table 
by  12  inches,  the  length  of 
the  specimen.  Then  the 
coefficient  of  elasticity  can 
be  computed  for  different 
values  of  5  and  s.  Thus  for  the  fourth  and  seventh  cases, 


Total 
Stress 
in 
Pounds. 

Stress 
per 
Square 
Inch. 

Elongation. 

Load 
on. 

Load 
off. 

2245 

5  OOO 

.OOI 

.OOO 

4490 

IO  OOO 

.004 

.OOO 

6735 

15  ooo 

.005 

.OOO 

8  980 

20000 

.008 

.OOO 

9878 

22  000 

.009 

.OOO 

10776 

24  ooo 

.OIO 

.OOO 

n  674 

26  ooo 

.OIO5 

.OOO 

12  572 

28  ooo 

.Oil 

.OOO 

13470 

30000 

.013 

.000 

14368 

32  ooo 

.014 

.000 

15  266 

34000 

.015 

.002 

16  164 

36  ooo 

.022 

.007 

17  062 

38  ooo 

.416 

•3995 

17  960 

40  ooo 

•  5445 

•523 

25  450 

50  ooo 

1.740 

1.707 

.23  175 

51  600 

2.468 

Specimen  broke  with  51  600  pounds 

per  square  inch. 

Stretch  in  12  inches,  2.468  inches. 

Stretch  in  8  inches,  1.812  inches. 

Stretch  in  8  inches,  22.65  Per  cent. 

Fractured  area,  0.297  square  inches. 

for  S  =  20  ooo,    s  = 


0.008 

12 


..      „        ,                0.0105 
for  S  =  26  ooo,    s  = 


and     E  =  30  ooo  ooo ; 
and    E  =  29  700  ooo. 


12  RESISTANCE  AND  ELASTICITY   OF   MATERIALS.       CH.  I. 

The  elastic  limit  was  reached  at  about  33  ooo  pounds  per 
square  inch,  indicated  by  the  beginning  of  the  set  and  the  rapid 
increase  of  the  elongations.  The  ultimate  tensile  strength  of 
the  specimen  was  5 1  600  pounds  per  square  inch.  The  ultimate 
unit-elongation  in  8  inches  of  the  length  was  0.226  inches  per 
linear  inch.  It  hence  appears  that  this  bar  of  wrought  iron  was 
higher  than  the  average  as  regards  stiffness,  elastic  limit  and 
ductility,  and  lower  than  the  average  in  ultimate  strength. 

The  '  working  strength '  of  a  material  is  that  unit-stress  to 
which  it  is,  or  is  to  be,  subjected.  This  should  not  be  greater 
than  the  elastic  limit  of  the  material,  since  if  that  limit  be  ex- 
ceeded there  is  a  permanent  set  which  impairs  the  elasticity. 
In  order  to  secure  an  ample  margin  of  safety  it  is  customary  to 
take  the  working  strength  at  from  one-third  to  two-thirds  the 
elastic  limit  Se.  The  reasons  which  govern  the  selection  of 
proper  values  of  the  working  strength  will  be  set  forth  in  the 
following  articles. 

To  investigate  the  security  of  a  piece  subjected  to  a  tension 
P,  it  is  necessary  first  to  divide  P  by  the  area  of  the  cross-sec- 
tion and  thus  determine  the  working  strength.  Then  a  com- 
parison of  this  value  with  the  value  of  Se  for  the  given  material 
will  indicate  whether  the  applied  stress  is  too  great  or  whether 
the  piece  has  a  margin  of  safety.  For  example,  if  a  tensile 
stress  of  4  500  pounds  be  applied  to  a  wrought  iron  bar  of  £ 
inches  diameter  the  working  unit-stress  is, 

S  =  —:-=  —    —  =  10  ooo.  pounds  per  square  inch,  nearly. 
Si       0.449 

As  this  is  less  than  one-half  the  elastic  limit  of  wrought  iron  the 
bar  has  a  good  margin  of  security. 

To  design  a  piece  to  carry  a  given  tension  P  it  is  necessary 
to  assume  the  kind  of  material  to  be  used  and  its  allowable 

p 
working  strength  S.     Then  -=  is  the  area  of  the  cross-section 


ART.  6.  COMPRESSION.  13 

of  the  piece,  which  may  be  made  of  such  shape  as  the  circum- 
stances of  the  case  require.  For  example,  if  it  be  required  to 
design  a  wooden  bar  to  carry  a  tensile  stress  of  4  500  pounds, 
the  working  strength  may  be  assumed  at  I  ooo  pounds  per 
square  inch  and  the  required  area  is  4.5  square  inches,  so  that 
the  bar  may  be  made  2  X  2j  inches  in  section. 

The  elongation  of  a  bar  within  the  elastic  limit  may  be  com- 
puted by  the  help  of  formula  (2).  For  instance,  let  it  be  re- 
quired to  find  the  elongation  of  a  wooden  bar  3X3  inches  and 
12  feet  long  under  a  tensile  stress  of  9  ooo  pounds.  From  the 
formulas  (2)  and  (i), 

E-S-P    •*•       •    K-Pl 

~  7  ~  'A  ^  /  '  ~  AE 

Substituting  in  this  the  values  E  =  I  500000,  A  =  9,  /—  144, 
and  P  =  9  ooo,  the  probable  value  of  the  elongation  X  is  found 
to  be  0.096  inches. 

Prob.  9.  Find  the  size  of  a  round  wrought  iron  rod  to  safely 
carry  a  tensile  stress  of  100  ooo  pounds. 

Prob.  10.  Compute  the  elongation  of  a  wooden  and  of  a  cast 
iron  bar,  each  being  2X3  inches  and  16  feet  long,  under  a  ten- 
sile stress  of  6  ooo  pounds. 

ART.  6.    COMPRESSION. 

The  phenomena  of  compression  are  similar  to  those  of  ten- 
sion, provided  that  the  length  of  the  specimen  does  not  exceed 
about  five  times  its  least  diameter.  The  piece  at  first  shortens 
proportionally  to  the  applied  stress,  but  after  the  elastic  limit  is 
passed  the  shortening  increases  more  rapidly,  and  is  accom- 
panied by  a  slight  enlargement  of  the  cross-section.  When  the 
stress  reaches  the  ultimate  strength  of  the  material  the  specimen 
cracks  and  ruptures.  If  the  length  of  the  piece  exceeds  about 
ten  times  its  least  diameter,  a  sidewise  bending  or  flexure  of 
the  specimen  occurs,  so  that  it  fails  under  different  circum- 


RESISTANCE   AND   ELASTICITY    OF   MATERIALS.       ClI.  I. 


stances  than  those  of  direct  compression.  All  the  values  given 
in  this  article  refer  to  specimens  whose  lengths  do  not  exceed 
about  five  times  their  least  diameter.  Longer  pieces  will  be 
discussed  in  Chapter  V  under  the  head  of  'columns.'  Owing 
to  the  difficulty  of  making  experiments  on  short  specimens,  the 
phenomena  of  compression  are  not  usually  so  regular  as  those 
of  tension. 

The  constants  of  compression  for  short  specimens  are  given 
in  the  following  table,  the  values,  like  those  for  tension,  being 
rough  average  values  liable  to  much  variation  in  particular 
cases. 


Material. 

Coefficient  of 
Elasticity,  E. 

Elastic 
Limit,  Se. 

Ultimate 
Compressive 
Strength,  Sc. 

Lbs.  per  sq.  in. 

Lbs.  per  sq.  in. 

Lbs.  per  sq.  in. 

Timber, 

I  500  000 

3  ooo 

8  000 

Brick, 

2  500 

Stone, 

6  ooo  ooo 

6  ooo 

Cast  Iron, 

15  ooo  ooo 

90000 

Wrought  Iron, 

25  ooo  ooo 

15  ooo 

55000 

Steel, 

30  ooo  ooo 

50  ooo 

150  ooo 

The  values  of  the  coefficient  of  elasticity  and  the  elastic  limit 
for  timber,  wrought  iron,  and  steel  here  stated  are  the  same  as 
those  for  tension,  but  the  same  reliance  cannot  be  placed  upon 
them,  owing  to  the  irregularity  of  experiments  thus  far  made. 
There  is  reason  to  believe  that  both  the  elastic  limit  and  the 
coefficient  of  elasticity  for  compression  are  somewhat  greater 
than  for  tension. 

The  investigation  of  a  piece  subjected  to  compression,  or  the 
design  of  a  short  piece  to  be  subjected  to  compression,  is 
effected  by  exactly  the  same  methods  as  for  tension.  Indeed 
it  is  customary  to  employ  these  methods  for  cases  where  the 
length  of  the  piece  is  as  great  as  ten  times  its  least  diameter. 


ART.  7. 


SHEAR. 


Prob.  ii.  Find  the  height  of  a  brick  tower  which  crushes 
under  its  own  weight.  Also  the  height  of  a  stone  tower. 

Prob.  12.  Compute  the  amount  of  shortening  in  a  wrought 
iron  specimen  i  inch  in  diameter  and  5  inches  long  under  a  load 
of  6000  pounds. 

ART.  7.    SHEAR. 

Shearing  stresses  and  strains  occur  whenever  two  forces,  act- 
ing like  a  pair  of  shears,  tend  to  cut  a  body  between  them. 
When  a  plate  is  punched  the  ultimate  shearing  strength  of  the 
material  must  be  overcome  over  the  surface  punched.  When 
a  bolt  is  in  tension  the  applied  stress  tends  to  shear  off  the 
head  and  also  to  strip  or  shear  the  threads  in  the  nut  and  screw. 
When  a  rivet  connects  two  plates  which  transmit  tension  the 
plates  tend  to  shear  the  rivet  across. 

The  ultimate  shearing  strength  of  materials  is  easily  deter- 
mined by  causing  rupture  under  a  stress  P,  and  then  dividing 
P  by  the  area  A  of  the  shorn  surface.  The  value  of  this  for 
timber  is  found  to  be  very  much  smaller  along  the  grain  than 
across  the  grain  ;  for  the  first  direction  it  is  sometimes  called 
longitudinal  shearing  strength  and  for  the  second  transverse 
shearing  strength.  The  same  distinction  is  sometimes  made 
in  rolled  wrought  iron  plates  and  bars  where  the  process  of 
manufacture  induces  a  more  or  less  fibrous  structure.  The 
elastic  limit  and  the  amount  of  detrusion  for  shearing  are  dif- 


Material. 

Coefficient  of 
Elasticity,  £. 

Ultimate 
Shearing 
Strength,^. 

Timber,  Longitudinal, 

4000CO 

600 

Timber,  Transverse, 

3000 

Cast  Iron, 

60OOOOO 

20000 

Wrought  Iron, 

15000000 

5OOOO 

Steel, 

7OOOO 

16  RESISTANCE  AND   ELASTICITY   OF   MATERIALS.        CH.  I. 

ficult  to  determine  experimentally.  The  coefficient  of  elas- 
ticity, however,  has  been  deduced  by  means  of  certain  calcula- 
tions and  experiments  on  the  twisting  of  shafts,  explained  in 
Chapter  VI  under  the  head  of  torsion. 

The  investigation  and  design  of  a  piece  to  withstand  shear- 
ing stress  is  made  by  means  of  the  equation  P  =  AS,  in 
the  same  manner  as  for  tension  and  compression.  As  an 

instance    of    investigation, 

v^x   *  •'''  ^ ^=*    wooden    specimen    shown 

Fig'2-  in  Fig.   2,  which  has   the 

following  dimensions :  length  ab  =  6  inches,  diameter  of  ends 
=  4  inches,  diameter  of  central  part  =  2.  inches.  Let  this 
specimen  be  subjected  to  a  tensile  stress  in  the  direction  of  its 
length.  This  not  only  tends  to  tear  it  apart  by  tension,  but 
also  to  shear  off  the  ends  on  a  surface  whose  length  is  ab  and 
whose  diameter  is  that  of  the  central  cylinder.  The  force  P 
required  to  cause  this  longitudinal  shearing  is, 

P  =  AS,  =  3.14  X2X6x6oo  =  22  600  pounds, 
while  the  force  required  to  rupture  the  specimen  by  tension  is, 

P=  ASt  =  3.14  X  i*  X  10  ooo  =  31  400  pounds. 
As  the  former  resistance  is  only  about  two-thirds  that  of  the 
•  latter  the  specimen  will  evidently  fail  by  the  shearing  off  of 
the  ends. 

When  a  bar  is  subject  either  to  tension  or  to  compression 
a  shear  occurs  in  any  section  except  those  perpendicular  and 

parallel  to  the  axis  of  the  bar. 
Let  Fig.  3  represent  a  bar  of 
cross-section  A  subject  to  the 
tensile  stress  P  which  produces 
Fig.  3.  in  every  section  perpendicular 

to  the  bar  the  unit-stress  —  .      Let  mn  be  a  plane  making  an 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.       17 

angle  0  with  the  axis,  and  cutting  from  the  bar  a  section  whose 
area  is  A1 .  On  the  left  of  the  plane  the  stress  P  may  be  resolved 
into  the  components  P,  and  P, ,  respectively  parallel  and  nor- 
mal to  the  plane,  and  the  same  may  be  done  on  the  right. 
Thus  it  is  seen  that  the  effect  of  the  tensile  stress  P  on  the 
plane  mn  is  to  produce  a  tension  P, -normal  to  it,  and  a  shear 
P,  along  it,  for  the  two  forces  P}  and  P1  act  in  parallel  planes 
and  in  opposite  directions.  The  shearing  stress  Pl  has  the 
value  P  cos  0,  which  is  distributed  over  the  area  Al  whose 
value  is  A  -~-  sin  8.  Hence  the  shearing  unit-stress  in  the 
given  section  is, 

P       P 

5,  =  —f-  =  -j  sin  0  cos  0. 

si  j         /i 

When  e  =  o°,  or  d  =  90°,  the  value  of  St  is  zero.  The  maxi- 

p 

mum  value  of  5,  occurs  when  0  =  45°,  and  then  5,  =  £— ,  or 

yi 

a  tensile  unit-stress  S  on  a  bar  produces  a  shearing  unit-stress 
of  £S  along  every  section  inclined  45  degrees  to  the  axis  of  the 
bar.  The  above  investigation  applies  also  to  compression  if 
the  direction  of  P  be  reversed,  and  it  is  sometimes  observed  in 
experiments  on  the  compression  of  short  specimens  that  rup- 
ture occurs  by  shearing  along  oblique  sections. 

Prob.  1 3.  A  hole  £  inches  in  diameter  is  punched  in  a  wrought 
iron  plate  f  inches  thick  by  a  pressure  on  the  punch  of  78  ooo 
pounds.  What  is  the  ultimate  shearing  strength  of  the  iron? 

Prob.  14.  A  wrought  iron  bolt  i£  inches  in  diameter  has  a 
head  f  inches  long.  Find  the  unit-stress  tending  to  shear  off 
the  head  when  a  tension  of  3  ooo  pounds  is  applied  to  the  bolt. 

ART.  8.    FACTORS  OF  SAFETY  AND  WORKING  STRESSES. 

The  factor  of  safety  for  a  body  under  stress  is  the  ratio  of  its 
ultimate  strength  to  the  actual  existing  unit-stress.  The  factor 
of  safety  for  a  piece  to  be  designed  is  the  ratio  of  the  ultimate 


IS 


RESISTANCE  AND  ELASTICITY  OF  MATERIALS.       CH.  I. 


strength  to  the  proper  allowable  working  strength.  Thus  if  St 
be  the  ultimate,  5  the  working  strength,  and  f  the  factor  of 
safety,  then 

/  =  — ' ,      and     St  =  fS. 

The  factor  of  safety  is  hence  always  an  abstract  number,  which 
indicates  the  number  of  times  the  working  stress  may  be  mul- 
tiplied before  the  rupture  of  the  body. 

The  law  (E)  in  Art.  3  indicates  that  working  stresses  should 
be  lower  for  shocks  and  sudden  stresses  than  for  steady  loads 
and  slowly  varying  stresses.  In  a  building  the  stresses  on  the 
walls  are  steady,  so  that  the  working  strength  may  be  taken 
high  and  hence  the  factor  of  safety  low.  In  a  bridge  the 
stresses  in  the  several  members  are  more  or  less  varying  in 
character  which  requires  a  lower  working  strength  and  hence 
a  higher  factor  of  safety.  In  a  machine  subject  to  shocks  the 
working  strength  should  be  lower  still  and  the  factor  of  safety 
very  high.  The  law  (E)  from  which  these  conclusions  are  de- 
rived is  not  merely  the  result  of  experience,  but  can  be  con- 
firmed by  theoretical  discussion  (Art.  80). 

The  following  are  average  values  of  the  allowable  factors  of 
safety  commonly  employed  in  American  practice.  These  values 


Material. 

For  Steady 
Stress. 
(Buildings.) 

For  Varying 
Stress. 
(Bridges.) 

For  Shocks. 
(Machines.) 

Timber, 

8 

1(5 

15 

Brick  and  Stone, 

15 

25 

30 

Cast  Iron, 

6 

15 

2O 

Wrought  Iron, 

4 

6 

10 

Steel, 

5 

7 

15 

are  subject  to  considerable  variation  in  particular  instances,  not 
only  on  account  of  the  different  qualities  and  grades  of  the 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.      19 

material,  but  also  on  account  of  the  varying  judgment  of 
designers.  They  will  also  vary  with  the  range  of  varying  stress, 
so  that  different  parts  of  a  bridge  may  have  very  different 
factors  of  safety. 

The  proper  allowable  working  strength  of  any  material  for 
tension,  compression,  or  shearing,  may  be  at  once  found  by 
dividing  the  ultimate  strength  by  the  proper  factor  of  safety. 
Regard  should  also  be  paid  to  the  elastic  limit  in  selecting  the 
working  strength,  particularly  for  materials  whose  elastic  limit 
is  well  defined.  For  wrought  iron  and  steel  the  working 
strength  should  be  well  within  the  elastic  limit,  as  already  in- 
dicated in  previous  articles.  For  cast  iron,  stone,  brick,  and 
timber  it  is  often  difficult  to  determine  the  elastic  limit,  and 
experience  alone  can  guide  the  proper  selection  of  the  working 
strength.  The  above  factors  of  safety  indicate  indeed  the  con- 
clusions of  experiment  and  experience  extending  over  the  past 
hundred  years. 

The  student  should  clearly  understand  that  the  exact  values 
given  in  this  and  the  preceding  articles  would  not  be  arbitrarily 
used  in  any  particular  case  of  design.  For  instance,  if  a  given 
lot  of  wrought  iron  is  to  be  used  in  an  engineering  structure, 
specimens  of  it  should  be  tested  to  determine  its  coefficient  of 
elasticity,  elastic  limit,  ultimate  strength,  and  percentage  of 
elongation.  Then  the  engineer  will  decide  upon  the  proper 
working  strength,  being  governed  by  its  qualities  as  shown  by 
the  tests,  the  character  of  the  stresses  that  come  upon  it,  and 
the  cost  of  workmanship. 

The  two  fundamental  principles  of  engineering  design  are 
stability  and  economy,  or  in  other  words : 

First,  the  structure  must  safely  withstand  all  the  stresses 

which  are  to  be  applied  to  it. 
Second,  the  structure  must  be  built  and  maintained  at  the 

lowest  possible  cost. 


2O  RESISTANCE   AND   ELASTICITY   OF   MATERIALS.       CH.  1. 

The  second  of  these  fundamental  principles  requires  that  all 
parts  of  the  structure  should  be  of  equal  strength,  like  the 
celebrated  'one-hoss  shay'  of  the  poet.  For,  if  one  part  is 
stronger  than  another,  it  has  an  excess  of  material  which  might 
have  been  spared.  Of  course  this  rule  is  to  be  violated  if  the 
cost  of  the  labor  required  to  save  the  material  be  greater  than 
that  of  the  material  itself.  Thus  it  often  happens  that  some 
parts  of  a  structure  have  higher  factors  of  safety  than  others, 
but  the  lowest  factors  should  not,  as  a  rule,  be  less  than  the 
values  given  above.  For  the  design  of  important  structures 
specifications  are  prepared  which  state  the  lowest  allowable 
unit-stresses  that  can  be  used. 

The  factors  of  safety  stated  above  are  supposed  to  be  so 
arranged  that,  if  different  materials  be  united,  the  stability  of 
all  parts  of  the  structure  will  be  the  same,  so  that  if  rupture 
occurs,  everything  would  break  at  once.  Or,  in  other  words, 
timber  with  a  factor  of  safety  8  has  about  the  same  reliability 
as  wrought  iron  with  a  factor  of  4  or  stone  with  a  factor  of  15, 
provided  the  stresses  are  due  to  steady  loads. 

The  assignment  of  working  strengths  with  regard  to  the 
elastic  limits  of  materials  is  more  rational  than  that  by  means 
of  the  factors  of  safety,  and  in  time  it  may  become  the  more 
important  and  valuable  method.  But  at  present  the  ultimate 
strengths  are  so  much  better  known  and  so  much  more  definitely 
determinable  than  the  elastic  limits  that  the  empirical  method 
of  factors  of  safety  seems  the  more  important  for  the  use  of 
students,  due  regard  being  paid  to  considerations  of  stiffness, 
elastic  limit,  and  ductility. 

As  an  example,  let  it  be  required  to  find  the  proper  size  of  a 
wrought  iron  rod  to  carry  a  steady  tensile  stress  of  90000 
pounds.  In  the  absence  of  knowledge  regarding  the  quality 
of  the  wrought  iron,  the  ultimate  strength  St  is  to  be  taken  as 


ART.  8.  FACTORS  OF  SAFETY  AND  WORKING  STRESSES.      21 

the  average  value,  55  ooo  pounds  per  square  inch.     Then,  for  a 
factor  of  safety  of  4,  the  working  strength  is, 

6"  = =  13  750  pounds  per  square  inch. 

4 

The  area  of  cross-section  required  is  hence, 

90  ooo 
A  = =  6.6  square  inches, 

which  may  be  supplied  by  a  rod  of  2^£  inches  diameter. 

Prob.  15.  Determine  the  size  of  a  short  steel  piston  rod  when 
the  piston  is  20  inches  in  diameter  and  the  steam  pressure  upon 
it  is  67.5  pounds  per  square  inch. 

Prob.  16.  A  wooden  frame  ABC  forming  an  equilateral  tri- 
angle consists  of  short  pieces  2X2  inches  jointed  at  A,  B,  and 
C.  It  is  placed  in  a  vertical  plane  and  supported  at  B  and  C 
so  that  BC  is  horizontal.  Find  the  unit-stress  and  factor  of 
safety  in  each  of  the  three  pieces  when  a  load  of  6  ooo  pounds 
is  applied  at  A. 


22  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.       CH.  II. 


CHAPTER   II. 
PIPES,  CYLINDERS,   AND   RIVETED  JOINTS. 

ART.  9.    WATER  AND  STEAM  PIPES. 

The  pressure  of  water  or  steam  in  a  pipe  is  exerted  in  every 
direction,  and  tends  to  tear  the  pipe  apart  longitudinally.  This 
is  resisted  by  the  internal  tensile  stresses  of  the  material.  If 
/  be  the  pressure  per  square  inch  of  the  water  or  steam,  d  the 
diameter  of  the  pipe  and  /  its  length,  the  force  P  which  tends 
to  cause  longitudinal  rupture  is  p .  Id.  This  is  evident  from  the 
fundamental  principle  of  hydrostatics  that  the  pressure  of  water 
in  any  direction  is  equal  to  the  pressure  on  a  plane  perpen- 
dicular to  that  direction,  or  may  be  seen  by  imagining  the  pipe 
to  be  filled  with  a  solid  substance  on  one  side  of  the  diameter, 
which  would  receive  the  pressure/  on  each  square  inch  of  the 
area  £/and  transmit  it  into  the  pipe.  If  t  be  the  thickness  of 
the  pipe  and  5  the  tensile  stress  which  is  uniformly  distributed 
over  it,  as  will  be  the  case  when  t  is  not  large  compared  with 
d,  the  resistance  on  each  side  is  // .  S.  As  the  resistance  must 
equal  the  pressure, 

pld  =  2tlS,       or      pd  =  2/5, 
which  is  the  formula  for  discussing  pipes  under  internal  pressure. 

The  unit-pressure/  for  water  may  be  computed  from  a  given 
head  h  by  rinding  the  weight  of  a  column  of  water  one  inch 
square  and  h  inches  high.  Or  if  h  be  giVen  in  feet,  the  pressure 
in  pounds  per  square  inch  may  be  computed  from  /  =  0.434/4. 

Water  pipes  maybe  made  of  cast  or  wrought  iron,  the  former 
being  more  common,  while  for  steam  the  latter  is  preferable. 


ART.  9.  WATER  AND   STEAM   PIPES.  2$ 

Wrought  iron  pipes  are  sometimes  made  of  plates  riveted  to- 
gether, but  the  discussion  of  these  is  reserved  for  another 
article.  A  water  pipe  subjected  to  the  shock  of  water  ram 
needs  a  high  factor  of  safety,  and  in  a  steam  pipe  the  factors 
should  also  be  high,  owing  to  shocks  liable  to  occur  from  con- 
densation and  expansion  of  the  steam.  The  formula  above 
deduced  shows  that  the  thickness  of  a  pipe  must  increase 
directly  as  its  diameter,  the  internal  pressure  being  constant. 

For  example,  let  it  be  required  to  find  the  factor  of  safety 
for  a  cast  iron  water  pipe  of  12  inches  diameter  and  -|  inches 
thickness  under  a  head  of  300  feet.  Here  p,  the  pressure  per 
square  inch,  equals  130.2  pounds.  Then  from  the  formula  the 
unit-stress  is, 

pd      130.2  X  12 
5  =  ^—  -  =  -  —  =  —  =  i  250  pounds  per  square  inch, 

2*  2  X  -g- 

and  hence  the  factor  of  safety  is, 
20000 


which  indicates  ample  security  under  ordinary  conditions. 

Again  let  it  be  required  to  find  the  proper  thickness  for  a 
wrought  iron  steam  pipe  of  18  inches  diameter  to  resist  a  pres- 
sure of  1  20  pounds  per  square  inch.  With  a  factor  of  safety 
of  10  the  working  strength  5  is  about  5  500  pounds  per  square 
inch.  Then  from  the  formula, 

pd      120  X  18  .     , 

t  =  !—  =  --  =  0.2  inches. 
2S      2x5  5oo 

In  order  to  safely  resist  the  stresses  and  shocks  liable  to  occur 
in  handling  the  pipes,  the  thickness  is  often  made  somewhat 
greater  than  the  formula  requires. 

Prob.  17.  What  should  be  the  thickness  of  a  cast  iron  pipe 
of  1  8  inches  diameter  under  a  head  of  300  feet? 

Prob.  1  8.  A  wrought  iron  pipe  is  3  inches  in  internal  diame- 


24  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.        CH.  II. 

ter  and  weighs  24  pounds  per  linear  yard.     What  steam  pres- 
sure can  it  carry  with  a  factor  of  safety  of  8  ? 


ART.  10.    THIN  CYLINDERS  AND  SPHERES. 

A  cylinder  subject  to  the  interior  pressure  of  water  or  steam 
tends  to  fail  longitudinally  exactly  like  a  pipe.  The  head  of 
the  cylinder  however  undergoes  a  pressure  which  tends  to 
separate  it  from  the  walls.  If  d  be  the  diameter  of  the  cylin- 
der and  /  the  internal  pressure  per  square  unit,  the  total  pres- 
sure on  the  head  is  ^nd*  .p.  US  be  the  working  unit-stress 
and  t  the  thickness  of  the  cylinder,  the  resistance  to  the  pres- 
sure is  approximately  itdtS,  if  /  be  so  small  that  5  is  uniformly 
dtstributed.  Since  the  resistance  must  equal  the  pressure, 

%nd* .  /  =  ndt  .  5,         or        pd  =  tfS. 

By  comparing  this  with  the  formula  of  the  last  article  it  is  seen 
that  the  resistance  of  a  pipe  to  transverse  rupture  is  double  the 
resistance  to  longitudinal  rupture. 

A  thin  sphere  subject  to  interior  pressure  tends  to  rupture 
around  a  great  circle,  and  it  is  easy  to  see  that  the  conditions 
are  exactly  the  same  as  for  the  transverse  rupture  of  a  cylin- 
der, or  that  pd  =  4/S.  For  thick  spheres  and  cylinders  the 
formulas  of  this  and  the  last  article  are  only  approximate. 

A  cylinder  under  exterior  pressure  is  theoretically  in  a  simi- 
lar condition  to  one  under  interior  pressure  as  long  as  it  re- 
mains a  true  circle  in  cross-section.  A  uniform  interior  pres- 
sure tends  to  preserve  and  maintain  the  circular  form  of  the 
cylindrical  annulus,  but  an  exterior  pressure  tends  at  once  to 
increase  the  slightest  variation  from  the  circle  and  render  it 
elliptical.  The  distortion  when  once  begun  rapidly  increases, 
and  failure  occurs  by  the  collapsing  of  the  tube  rather  than  by 
the  crushing  of  the  material.  The  flues  of  a  steam  boiler  are 
the  most  common  instance  of  cylinders  subjected  to  exterior 


ART.  10. 


THIN   CYLINDERS   AND   SPHERES. 


pressure.  In  the  absence  of  a  rational  method  of  investigating 
such  cases  recourse  has  been  had  to  experiment.  Tubes  of 
various  diameters,  lengths,  and  thicknesses  have  been  subjected 
to  exterior  pressure  until  they  collapse  and  the  results  have 
been  compared  and  discussed.  The  following  for  instance  are 
the  results  of  three  experiments  by  FAIRBAIRN  on  wrought 
iron  tubes. 


Length 
in 
Inches. 

Diameter, 
in 
Inches. 

Thickness 
in 
Inches. 

Pressure 
per 
Sq.  Inch. 

37 

9 

0.14 

378 

60 

14* 

0.125 

"5 

6l 

i8f 

0.25 

420 

From  these  and  other  similar  experiments  it  has  been  concluded 
that  the  collapsing  pressure  varies  directly  as  some  power  of  the 
thickness,  and  inversely  as  the  length  and  diameter  of  the  tube. 
For  wrought  iron  tubes  WOOD  gives  the  empirical  formula  for 
the  collapsing  pressure  per  square  inch, 

^  =  9600000^-. 

The  values  of  p  computed  from  this  formula  for  the  above 
three  experiments  are  397,  120,  and  409,  which  agree  well  with 
the  observed  values. 

The  proper  thickness  of  a  wrought  iron  tube  to  resist  ex- 
terior pressure  may  be  readily  found  from  this  formula  after 
assuming  a  suitable  factor  of  safety.  For  example,  let  it  be 
required  to  find  t  when  p  =  120  pounds  per  square  inch,  /=  72 
inches,  d  •=  4  inches  and  the  factor  of  safety  =  10.  Then 

,,I8  =  io  X  120  X  72  X_4  =  Q     6 
9600000 

from  which  with  the  help  of  logarithms  the  value  of  t  is  found 
to  be  0.22  inches. 


26  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.       CH.  II. 

Prob.  19.  What  interior  pressure  per  square  inch  will  burst 
a  cast  iron  sphere  of  24  inches  diameter  and  f  inches  thickness. 

Prob.  20.  What  exterior  pressure  per  square  inch  will  col- 
lapse a  wrought  iron  tube  72  inches  long,  4  inches  diameter  and 
0.25  inches  thickness? 

ART.  ir.    THICK  CYLINDERS. 

When  the  walls  of  a  cylinder  are  thick  compared  with  its 
interior  diameter  it  cannot  be  supposed,  as  in  the  preceding 
articles,  that  the  stress  is  uniformly  distributed  over  the  thick- 
ness t.     Let  Fig.  4  represent  one-half  of  a  section  of  a  thick 
cylinder  subject  to  interior  pressure 
over  the  length  /,  tending  to  produce 
longitudinal  rupture.     Let  r  and  rl 
be   the   interior  and   exterior   radii, 
then  rv  —  r  =  t  the  thickness.     Let 
5  and  5,  be  the  tensile  unit-stresses 
at  the  inner  and  outer  edges  of  the 
Fi*-4-  annulus.     Before  the  application  of 

the  pressure  the  volume  of  the  annulus  is  n(r*  —  r2)/,  after  the 
pressure  is  applied  the  radius  rl  is  increased  to  r,  -j-  Ji  and  r  to 
r  -j-  y,  so  that  its  volume  is  7t(rl  -|-  y^fl  —  n(r  -j-  y}*l.  The  an- 
nulus is  however  really  changed  only  in  form,  so  that  the  two 
expressions  for  the  volume  are  equal,  and  equating  them  gives, 


or,  since  y  and  yl  are  small  compared  with  r  and  rl  their 
squares  may  be  neglected,  and  hence 

y      r-> 
ry  =  rlyl  ,         or         ^-  =  -  . 

Now  if  the  material  is  not  stressed  beyond  the  elastic  limit  the 
unit-stresses  5"  and  Sl  are  proportional  to  the  corresponding 
unit-elongations.  The  elongation  of  the  inner  circumference 
is  2ity  and  that  of  the  outer  circumference  is  2rtyl  ,  and  divid- 


ART.  II.  THICK  CYLINDERS.  27 

ing  these  by  2nr  and  2nrl  respectively  the  unit-elongations  are 
found;  then, 


S,      r       r,       r     y 

Substituting  in  this  the  value  of  the  ratio  —  as  above  found, 
gives 


that  is,  the  unit-stresses  in  the  walls  of  the  cylinder  vary  in- 
versely as  the  squares  of  their  distances  from  the  center. 

The  total  stress  acting  over  the  area  2t  .  /  is  now  to  be  found 
by  summing  up  the  unit-stresses.  Let  Sx  be  any  unit-stress  at 
a  distance  x  from  the  center,  and  S,  as  before,  be  that  at  the 
inner  circumference,  which  is  the  greatest  of  all  the  unit-stresses. 
Then  by  the  law  of  variation, 


The  stress  acting  over  the  area  /.  dx  is  then 


and  the  total  stress  over  the  area  2t  .  /  is 


r  +  t 

This  is  the  value  of  the  internal  resisting  stress  in  the  walls  of 
the  pipe  ;  if  /  be  neglected  in  comparison  with  r  it  reduces  to 
2Slt  which  is  the  same  as  previously  found  for  thin  cylinders ; 
if  t  =  r  it  becomes  Sit  or  only  one-half  the  resistance  of  a  thin 
cylinder. 

The  total  interior  pressure  which  tends  to  rupture  the  cylin- 
der longitudinally  is  2rl .  p,  if  p  be  the  unit-pressure  (Art.  9). 


28  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.       CH.  II. 

Equating  this  to  the  total  internal  resisting-stress  gives  the 
formula, 


from  which  one  of  the  quantities  S,  p,  r,  or  t  can  be  computed 
when  the  other  three  are  given.  For  instance,  let  this  be  ap- 
plied to  the  same  example  as  in  Art.  9,  where/  =  130.2  pounds 
per  square  inch,  2r  =  12  inches,  /  =  f  inch,  and  the  value  of  5 
is  required,  then 

S  =  —  -f  p  =  1250+  130  =  1380  pounds  per  square  inch, 

which  is  about  10  per  cent  greater  than  the  value  found  by  the 
approximate  formula  for  thin  pipes. 

Prob.  21.  Prove  when  the  thickness  of  a  pipe  equals  its  in- 
terior radius  that  the  exterior  circumference  elongates  one-half 
as  much  as  the  interior  circumference. 

Prob.  22.  If.  a  gun  of  3  inches  bore  is  subject  to  an  interior 
pressure  of  I  8oc  pounds  per  square  inch,  what  should  be  its 
thickness  so  that  the.  greatest  stress  on  the  material  may  not 
exceed  3  ooo  pounds  per  square  inch  ? 

ART.  12.    INVESTIGATION  OF  RIVETED  JOINTS. 

When  two  plates  which  are  under  tension  are  joined  together 
by  rivets,  these  must  transfer  that  tension  from  one  plate  to 
another.  A  shearing  stress  is  thus  brought  upon  each  rivet 
which  tends  to  cut  it  off.  A  compressive  stress  is  also  brought 
sidewise  upon  each  rivet  which  tends  to  crush  it  ;  this  particu- 
lar kind  of  compression  is  often  called  "bearing  stress."  The 
exact  manner  in  which  it  acts  upon  the  cylindrical  surface  of 
the  rivet  is  not  known,  but  it  is  usually  supposed  to  be  equiva- 
lent to  a  stress  uniformly  distributed  over  the  projection  of  the 
surface  on  a  plane  through  the  axis  of  the  rivet. 

Case  I.  Lap  Joint  with  single  riveting.  —  Let  P  be  the  tensile 
stress  which  is  transmitted  from  one  plate  to  the  other  by 


ART.   12.         INVESTIGATION   OF   RIVETED   JOINTS. 


29 


means   of  a  single   rivet,  t  the  thickness   of  the  plates,  d  the 

diameter  of  a  rivet,  and  a  the  pitch  of  the  rivets.     Let  St,  5,, 

and  St  be    the    unit-stresses   in 

tension,  shear,  and  compression 

produced  by  P  upon  the  plates 

and  rivets.     Then   for  the  ten- 

sion on  the  plate, 


for  the  shear  on  the  rivet, 


and  for  compression  on  the  rivet, 

Fig.  5- 

P  =  td.Se. 

From  these  equations  the  unit-stresses  may  be  computed,  when 
the  other  quantities  are  known,  and  by  comparing  them  with 
the  proper  working  unit-stresses  the  degree  of  security  of  the 
joint  is  estimated. 

Case  II.  Lap  Joint  with  double  riveting. — In  this  arrange- 
ment  the   plates    have   a  wider   lap,     /^"X — "v 

and    there    are    two    rows    of    rivets.  ' — ^ —                        ^  -\ 
Let  a  be  the  pitch    of  the  rivets  in 
one  row,  then  the  tensile  stress  P  is 
distributed  over  two   rivets,  and  the 
three  formulas  are, 

P=t(a-d}St, 


P=  2.tdSc, 

from  which  the  unit-stresses  may  be 
computed  and  the  strength  of  the 
joint  be  investigated.  The  loss  of  Fig.  6. 

strength  is  here  generally  less  than  in  the  previous  case  since  a 
can  be  made  larger  with  respect  to  d. 


30  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.       CH.  II. 

Case  III.  Butt  Joint  with  single  riveting. — For  this  arrange- 
^ — ^  ^ — >.  ment  the  shear  on  each  of 

tne    rivets    comes    on    two 
cross-sections,  which  is  said 
Fig.  7.  to  be  a  case  of  double  shear, 

and  the  formulas  are, 

P=t(a-  d)St,  =  2  .  ±nd*Sg  =  tdSe. 

Accordingly,  a  lap  joint  with  double  riveting  has  the  same  ten- 
sile and  shearing  strength  as  a  butt  joint  with  single  riveting, 
if  the  values  of  «,  d,  and  t  be  equal  in  the  two  cases ;  the  bear- 
ing resistance,  however,  is  only  one  half  as  large. 

For  example,  let  it  be  required  to  investigate  a  single  riveted 
butt  joint  consisting  of  plates  0.75  inches  thick  with  covers 
0.375  inches  thick,  and  rivets  of  I  inch  diameter  and  4  inches 
pitch,  when  a  tension  of  8  ooo  pounds  is  transmitted  through 
one  rivet.  First,  the  working  tensile  unit-stress  on  the  plate  is 
found  to  have  the  value, 

8000 
St  =  — — =  3  560  pounds  per  square  inch. 

Next  the  shearing  unit-stress  on  the  rivets  is, 

8000 
os  =  — — x—  =  5  100  pounds  per  square  inch. 

2  X  0.7^5 

Lastly,  the  bearing  compressive  unit-stress  on  the  rivets  is, 

8000 
Sc  = =  10  700  pounds  per  square  inch. 

It  thus  appears  that  the  joint  has  the  greatest  factor  of  safety 
against  tension  and  the  least  against  compression  of  the  rivets. 
It  should  be  said,  however,  that  for  wrought  iron  plates  and 
rivets  the  highest  allowable  working  stresses  for  tension,  shear, 
and  bearing  are  generally  considered  to  be  about  9  ooo,  7  500, 
and  1 2  ooo  pounds  per  square  inch  respectively;  hence  the 


ART.  12.         INVESTIGATION  OF  RIVETED  JOINTS.     ,  31 

joint  has  proper  security  under  the  given  conditions  although 
the  degree  of  security  is  quite  different  for  the  different 
stresses. 

The  '  efficiency  '  of  a  joint  is  defined  to  be  the  ratio  of  its 
highest  allowable  stress  to  the  highest  allowable  stress  of  the 
unriveted  plate.  The  highest  allowable  stresses  in  tension, 
shear,  and  compression  are  the  three  expressions  for  P,  using, 
for  wrought  iron,  the  values  above  mentioned  ;  and  the  highest 
allowable  stress  of  the  unriveted  plate  is  atSt  .  Thus  result  the 
following  values  of  the  efficiency, 

a  —  d 

For  tension,  e  =  -  » 


For  shear, 


m  .  dSc 
For  compression,  e  =  —  ^r—  , 

in  which  m  denotes  the  number  of  rivets  in  the  width  a  which 
transmit  the  tension  P  and  n  denotes  the  number  of  rivet-sec- 
tions in  the  same  space  over  which  the  shear  is  distributed. 
The  smallest  of  these  values  of  e  is  to  be  taken  as  the  efficiency 
of  the  joint.  Thus  for  the  above  numerical  example  the  three 
values  are  0.75,  0.44,  and  0.33  ;  the  working  strength  of  the 
joint  is,  hence,  only  33  per  cent  of  that  of  the  unriveted  plate. 

If  in  the  above  formulas,  St,  Ss,  and  Sc  be  taken  as  the  ulti- 
mate strengths,  the  resulting  values  of  e  will  be  the  efficiencies 
at  the  moment  of  rupture  of  the  joint.  For  the  same  numerical 
example  the  three  ultimate  efficiencies  are  0.75,  0.48,  and  0.25. 

Prob.  23.  A  boiler  is  to  be  formed  of  wrought  iron  plates  f 
inches  thick,  united  by  single  lap  joints  with  rivets  f  inches 
diameter  and  if  inches  pitch.  Find  the  efficiency  of  the  joint. 
Find  the  factor  of  safety  of  the  boiler  if  it  is  30  inches  in  di- 
ameter and  carries  a  steam  pressure  of  100  pounds. 


32  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.       CH.  II. 

ART.  13.  DESIGN  OF  RIVETED  JOINTS. 
A  theoretically  perfect  joint  with  regard  to  strength  is  one 
so  arranged  that  all  parts  (like  the  one-hoss  shay)  have  the 
same  degree  of  security.  Thus  the  resistance  of  the  plate  to 
tension  must  equal  the  resistance  of  the  rivets  to  shearing,  and 
each  of  these  must  equal  the  resistance  of  the  rivets  to  com- 
pression. The  three  expressions  for  P  of  the  last  Article  should 
hence  be  equal,  or,  what  amounts  to  the  same  thing,  the  three 
efficiencies  should  be  equal.  Equating  then  the  second  to  the 
third  and  solving  for  d,  gives 


from  which  d  can  be  computed  when  /  is  assumed.     Again, 
equating  the  first  and  third  and  eliminating  d  gives, 


AfinSc  I         mSc\ 

from  which  the  pitch  of  the  rivets  can  be  obtained.  Inserting 
these  values  of  d  and  a  in  either  of  the  expressions  for  e  fur- 
nishes the  formula, 

'~«+j5' 

from  which  the  efficiency  can  be  ascertained.  The  best  joint 
will  be  that  which  has  the  least  loss  of  strength  due  to  the 
riveting,  or  that  which  has  a  value  of  e  as  near  unity  as  possible. 

Using  for  wrought  iron  plates  and  rivets  the  working  unit- 
stresses  St  =  9000,  5S  =  7  500,  and  Sc  —  12000  pounds  per 
square'  inch,  the  above  formulas  for  a  lap  joint  with  single  row 
of  rivets  where  m  —  i  and  n  =  i,  reduce  to, 

d  =  2.04/,  a  =  4.75^,  e  —  0.57, 

so  that,  if  the  thickness  of  the  plate  be  given  and  the  diameter 
and  pitch  of  the  rivet  be  made  according  to  these  ruks,  the 


ART.  13.  DESIGN   OF  RIVETED  JOINTS.  33 

joint  has  about  57  per  cent  of  the  strength  of  the  unholed 
plate.  For  a  lap  joint  with  double  riveting  where  m  =  2  and 
n  =  2,  they  become 

d  =  2.O4/,  a  —  7.48^,  e  =  0.73. 

This  example  shows  clearly  the  advantage  of  double  over  single 
riveting,  and  by  adding  a  third  row  the  efficiency  will  be  raised 
to  about  80  per  cent. 

The  application  of  the  above  formulas  to  butt  joints  makes 
the  diameter  of  the  rivet  equal  to  the  thickness  of  the  plate 
and  makes  the  pitch  much  smaller  than  the  above  values  for 
lap  joints.  These  proportions  are  difficult  to  apply  in  practice 
on  account  of  the  danger  of  injuring  the  metal  in  punching  the 
holes.  For  this  reason  joints  are  often  made  in  which  the 
strengths  of  the  different  parts  are  not  equal.  Many  other 
reasons,  such  as  cost  of  material  and  facility  of  workmanship, 
influence  also  the  design  of  a  joint  so  that  the  formulas  above 
deduced  are  to  be  regarded  only  as  a  rough  guide.  The  old 
rules  which  are  still  often  used  for  determining  the  pitch  in  butt 
joints,  are 


the  first  being  for  single  and  the  second  for  double  riveting. 
These  are  deduced  by  making  the  strength  of  the  joint  equal 
in  tension  and  shear,  and  taking  5S  =  St. 

It  may  be  required  to  arrange  a  joint  so  as  to  secure  eithei 
strength  or  tightness.  For  a  bridge,  strength  is  mainly  needed  ; 
for  a  gasholder,  tightness  is  the  principal  requisite;  while  for  a 
boiler  both  these  qualities  are  desirable.  In  general  a  tight 
joint  is  secured  by  using  small  rivets  with  a  small  pitch.  The 
lap  of  the  plates,  and  the  distance  between  the  rows  of  rivets, 
is  determined  by  practical  considerations  rather  than  by 
theoretic  formulas. 


34  PIPES,   CYLINDERS,   AND   RIVETED  JOINTS.        CH.  II. 

Prob.  24.  A  lap  joint  with  double  riveting  is  to  be  formed 
of  plates  £  inches  thick  with  rivets  f  inches  diameter.  Find 
the  pitch  so  that  the  "strength  of  the  plate  shall  equal  the 
shearing  strength  of  the  rivets,  and  compute  the  efficiency  of 
the  joint. 

ART.  14.    MISCELLANEOUS  EXERCISES. 

It  will  be  profitable  to  the  student  to  thoroughly  perform  the 
following  exercises  and  problems  and  to  write  upon  each  a 
detailed,  report,  which  should  contain  all  the  sketches  and 
computations  necessary  to  clearly  explain  the  data,  the  reason- 
ing, the  computations,  and  the  conclusions.  Problem  26  is 
intended  for  students  proficient  in  the  use  of  calculus. 

Exercise  I.  Visit  an  establishment  where  tensile  tests  are 
made.  Ascertain  the  kind  of  machine  employed,  its  capacity, 
the  method  of  applying  the  stresses,  the  method  of  measuring 
the  stresses,  the  method  of  measuring  the  elongations.  Ascer- 
tain the  kind  of  material  tested,  the  reason  for  testing  it,  and 
the  conclusions  derived  from  the  tests.  Give  full  data  for  the 
tests  on  four  different  specimens,  compute  the  values  of  co- 
efficient of  elasticity,  ultimate  strength  and  ultimate  elongation 
for  them,  and  state  your  conclusions. 

Exercise  2.  Procure  a  wrought  iron  bolt  and  nut.  Measure 
the  diameter  of  bolt,  length  of  head,  and  length  of  nut.  State 
the  equation  of  condition  that  the  head  of  the  bolt  may  shear 
off  at  the  same  time  the  bolt  ruptures  under  tension.  Com- 
pute the  length  of  head  for  a  given  diameter.  Explain  why 
the  length  of  the  head  is  made  greater  than  theory  apparently 
requires.  Compile  a  table  giving  dimensions  of  bolts  and  nuts 
of  different  diameters. 

Exercise  3.  Go  to  a  boiler  shop  and  witness  operations  upon  a 
boiler  in  process  of  construction.  Ascertain  length  and  diameter 
of  boiler,  thickness,  pitch  and  diameter  of  rivets,  method  of 
forming  holes,  method  of  doing  the  riveting.  Compute  the  loss 
of  strength  caused  by  the  riveting.  Compute  the  steam  pressure 


ART.  14.  MISCELLANEOUS   EXERCISES.  35 

which  would  cause  longitudinal  rupture  of  the  plate  along  a 
line  of  rivets.  Ascertain  whether  the  joint  is  proportioned  in 
accordance  with  theory. 

Prob.  25.  A  wrought  iron  pipe  f  inches  thick  and  20  inches 
in  diameter  is  to  be  subjected  to  a  head  of  water  of  345  feet. 
Compute  the  probable  increase  in  diameter  due  to  the  interior 
pressure,  regarding  the  pipe  as  thin. 

Prob.  26.  Let  a  pier  whose  top  width  is  b  and  length  /  support 
a  uniformly  distributed  load  P.  Let  the  width  of  the  pier  at  a 
distance  y  below  the  top  be  x,  its  constant  length  being  /  at  all 
horizontal  sections.  Let  w  be  the  weight  of  the  masonry  per 
cubic  foot.  Prove  that,  in  order  to  make  the  compressive 
unit-stress  the  same  for  all  horizontal  sections,  the  profile  of 

the  pier  must  be  such  as  to  satisfy  the  equation  y  =  —  log,,  -j- 

in  which  6"  =  -7-7. 
bl 


36  CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.        CH.  III. 


CHAPTER  III. 

CANTILEVER  BEAMS  AND  SIMPLE  BEAMS. 
ART.  15.    DEFINITIONS. 

Transverse  stress,  or  flexure,  occurs  when  a  bar  is  in  a 
horizontal  position  upon  one  or  more  supports.  The  weight 
of  the  bar  and  the  loads  upon  it  cause  it  to  bend  and  induce 
in  it  stresses  and  strains  of  a  complex  nature  which,  as  will  be 
seen  later,  may  be  resolved  into  those  of  tension,  compression, 
and  shear.  Such  a  bar  is  called  a  beam. 

A  simple  beam  is  a  bar  resting  upon  supports  at  its  ends. 
A  cantilever  beam  is  a  bar  on  one  support  in  its  middle,  or  the 
portion  of  any  beam  projecting  out  of  a  wall  or  beyond  a  sup- 
port may  be  called  a  cantilever  beam.  A  continuous  beam  is 
a  bar  resting  upon  more  than  two  supports.  In  this  book  the 
word  beam,  when  used  without  qualification,  includes  all  kinds, 
whatever  be  the  number  of  the  supports,  or  whether  the  ends 
be  free,  supported,  or  fixed. 

The  elastic  curve  is  the  curve  formed  by  a  beam  as  it  deflects 
downward  under  the  action  of  its  own  weight  and  of  the  loads 
upon  it.  Experience  teaches  that  the  amount  of  this  deflection 
and  curvature  is  very  sm£ll.  A  beam  is  said  to  be  fixed  at  one 
end  when  it  is  so  arranged  that  the  tangent  to  the  elastic  curve 
at  that  end  always  remains  horizontal.  This  may  be  done  in 
practice  by  firmly  building  one  end  into  a  wall.  A  beam  fixed 
at  one  end  and  unsupported  at  the  other  is  a  cantilever  beam. 

The  loads  on  beams  are  either  uniform  or  concentrated.  A 
uniform  load  embraces  the  weight  of  the  beam  itself  and  any 
load  evenly  spread  over  it.  Uniform  loads  are  estimated  by 
their  intensity  per  unit  of  length  of  the  beam,  and  usually  in 


ART    l6.  REACTIONS  OF  THE  SUPPORTS.  37 

pounds  per  linear  foot.  The  uniform  load  per  linear  unit  is 
designated  by  «/,  then  wx  will  represent  the  load  over  any 
distance  x.  If  /  be  the  length  of  the  beam,  the  total  uniform 
load  is  wl  which  may  be  represented  by  W.  A  concentrat- 
ed load  is  a  weight  applied  at  a  definite  point  and  is  desig- 
nated by  P. 

In  this  chapter  cantilever  and  simple  beams  will  be  princi- 
pally discussed,  although  all  the  fundamental  principles  and 
methods  hold  good  for  restrained  and  continuous  beams  as 
well.  Unless  otherwise  stated  the  beams  will  be  regarded  as 
of  uniform  cross-section  throughout,  and  in  computing  their 
weights  the  rules  of  Art.  I  will  be  found  of  service. 

Prob.  27.  Find  the  diameter  of  a  round  steel  bar  which 
weighs  48  pounds,  its  length  being  4  feet. 

ART.  16.  REACTIONS  OF  THE  SUPPORTS. 
The  points  upon  which  a  beam  is  supported  react  upward 
against  the  beam  an  amount  equal  to  the  pressure  of  the  beam 
upon  them.  The  beam,  being  at  rest,  is  a  body  in  equilibrium 
under  the  action  of  a  system  of  forces  which  consist  of  the 
downward  loads  and  the  upward  reactions.  The  loads  are 
usually  given  in  intensity  and  position,  and  it  is  required  to 
find  the  reactions.  This  is  effected  by  the  application  of  the 
fundamental  conditions  of  static  equilibrium,  which  for  a 
system  of  vertical  forces,  are, 

2  of  all  vertical  forces  =  o, 

2  of  moments  of  all  forces  =  o. 

The  first  of  these  conditions  says  that  the  sum  of  all  the 
loads  on  the  beam  is  equal  to  the  sum  of  the  reactions.  Hence 
if  there  be  but  one  support,  this  condition  gives  at  once  the 
reaction. 

For  two  supports  the  second  condition  must  be  used  in  con- 
nection with  the  first.  The  center  of  moments  may  be  taken 


38  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

anywhere  in  the  plane,  but  it  is  more  convenient  to  take  it  at  one 

of  the  supports.     For  example,  consider  a  single  concentrated 

load  P  situated  at  4  feet  from  the 

IP  left  end  of  a  simple  beam  whose 

4  span  is  13  feet.     The  equation  of 


r       moments,  with  the  center   at  the 

Fig.  8. 


I*  "*        left     support,    is     13^-4^  = 

from  which  R,  =  -feP.  Again  the 
equation  of  moments,  with  the  center  at  the  right  support,  is 
13^,  —  9/>=  o,  from  which  R^  —  ^jP-  As  a  check  it  may  be 
observed  that  R,  +  Rt  =  P. 

For  a  uniform  load  over  a  simple  beam  it  is  evident,  without 
applying  the  conditions  of  equilibrium,  that  each  reaction  is 
one-half  the  load. 

The  reactions  due  to  both  uniform  and  concentrated  loads 
on  a  simple  beam  may  be  obtained  by  adding  together  the 
reactions  due  to  the  uniform  load  and  each  concentrated  load, 
or  they  may  be  computed  in  one  operation.  As  an  example 
of  the  latter  method  let  Fig.  9  represent  a  simple  beam  12  feet 
in  length  between  the  supports  and  weighing  35  pounds  per 
.  linear  foot,  its  total  weight 

! 31^2^—3-'  ^  —4.'— 4      being    420     pounds.      Let 

L  =^     there  be  three  concentrated 

\KI  -Rsj      loads   of   300,  60,  and    150 

Fig- 9>  pounds  placed  at  3,  5,  and 

8  feet  respectively  from  the  left  support.     To  find  the  right 

reaction  R,  the  center  of  moments  is  taken  at  the  left  support, 

and  the  weight  of  the  beam  regarded  as  concentrated  at  its 

middle ;  then  the  equation  of  moments  is, 

R,  X  12  =  420  X  6  + 300  X  3  +  60  X  5  +  150  X  8 
from  which  ^,  =  410  pounds.     In  like  manner  to  find  R,  the 
center  of  moments  is  taken  at  the  right  support,  and 

R,  x  12  =  420  x  6  +  300x9  +  60  x  7+150x4 


ART.  17. 


THE   VERTICAL   SHEAR. 


39 


from  which  R^  =  520  pounds.  As  a  check  the  sum  of  Rl  and 
R9  is  seen  to  be  930  pounds  which  is  the  same  as  the  weight  of 
the  beam  and  the  three  loads. 

When  there  are  more  than  two  supports  the  problem  of  find- 
ing the  reactions  from  the  principles  of  statics  becomes  inde- 
terminate, since  two  conditions  of  equilibrium  are  only  sufficient 
to  determine  two  unknown  quantities.  By  introducing,  how- 
ever, the  elastic  properties  of  the  material,  the  reactions  of 
continuous  beams  may  be  deduced,  as  will  be  explained  in 
Chapter  IV. 

Prob.  28.  A  simple  beam  12  feet  long  weighs  20  pounds  per 
linear  foot  and  carries  a  load  of  500  pounds.  Where  should 
this  load  be  put  so  that  one  reaction  may  be  double  the  other  ? 

Prob.  29.  A  simple  beam  weighing  30  pounds  per  linear  foot 
is  1 8  feet  long.  A  weight  of  700  pounds  is  placed  5  feet  from 
the  left  end  and  one  of  500  pounds  at  8  feet  from  the  right 
end.  Find  the  reactions  due  to  the  total  load. 

ART.  17.    THE  VERTICAL  SHEAR. 

When  a  beam  is  short  it  sometimes  fails  by  shearing  in  a 
vertical  section  as  shown  in  Fig.  10.  The  external  force  which 
produces  this  shearing  on  any  section 
is  the  resultant  of  all  the  vertical  forces 
on  one  side  of  that  section.  Thus,  in 
the  second  diagram  the  resultant  of  all 
these  external  forces  is  the  loads  and 
the  weight  of  the  part  of  the  beam  on 
the  left  of  the  section  ;  in  the  third  dia- 
gram the  resultant  is  the  loads  and  the 
weight  on  the  right,  or  it  is  reaction  at  Fig.  10. 

the  wall  minus  the  weight  of  the  beam  between  the  wall  and 
the  section. 

'Vertical  Shear '  is  the  name  given  to  the  algebraic  sum  of 
all  external  forces  on  the  left  of  the  section  considered.  Let 


4O  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

it  be  denoted  by  V,  then  for  any  section  of  a  simple  or  can- 
tilever beam, 

V  =  Left  reaction  minus  all  loads  on  left  of  section. 
Here  upward  forces  are  regarded  as  positive  and  downward 
forces  as  negative.     V  is  hence  positive  or  negative  according 
as  the  left  reaction  exceeds  or  is  less  than  the  loads  on  the  left 
of  the  section.     To  illustrate,  consider  a  simple  beam  loaded 
in  any  manner  and  cut  at  any  sec- 
pi  Ft  I     Jp  tion  by  a  vertical  plane  mn.     Let 
I — * — l|t                              ~1  R1  be  the  left  and  R,  the  right  re- 
||F                         ?  action.     Let  2Pt  denote  the  sum 
-B'l     of  all  the  loads  on  the  left  of  the 
Fig-  "•  section  and  2P^  the  sum  of  those 
on  the  right.     Then,  from  the  definition, 

V-R,  -2P,. 

Since  R,  -f  R,  =  2P>  +  2P9  it  is  clear  if  R,  —  2P,  =  +V 
that  R,  —  2P,  =  —  V,  or  that  the  resultant  of  all  the  external 
forces  on  one  side  of  the  section  is  equal  and  opposite  to  the 
resultant  of  those  on  the  other  side.  They  form,  in  short,  a 
pair  of  shears  acting  on  opposite  sides  of  the  section  and  tend- 
ing to  cause  a  sliding  or  detrusion  along  the  section.  The 
value  of  the  vertical  shear  for  any  section  of  a  simple  beam  or 
cantilever  is  readily  found  by  the  above  equation.  When  R1 
exceeds  2Pt ,  the  vertical  shear  V  is  positive,  and  the  left  part 
of  the  beam  tends  to  slide  upward  relative  to  the  right  part. 
When  Rt  is  less  than  2P, ,  the  vertical  shear  V  is  negative, 
and  the  left  part  tends  to  slide  downward  relative  to  the  other. 
In  the  upper  diagram  of  Fig.  10  the  shear  in  the  left  hand  sec- 
tion is  positive  and  that  in  the  right  hand  section  is  negative. 

The  vertical  shear  varies  greatly  in  value  at  different  sections 
of  a  beam.  Consider  first  a  simple  beam  /  feet  long  and  weigh- 
ing w  pounds  per  linear  foot.  Each  reaction  is  then  %wl. 
Pass  a  plane  at  any  distance  x  from  the  left  support,  then  from 


ART.  17. 


THE   VERTICAL   SHEAR. 


the  definition  the  vertical  shear  for  that  section  is  V=^  \wl  —  wx. 
Here  it  is  seen  that  V  has  its  greatest  value  \wl  when  x  =  o, 

that    V   decreases   as  x  increases,      ^ ^ ^ 

and  that  F  becomes  o  when  x  =  \l.   f~  ~"1 

When   x   is   greater  than  £/,  F  is 

negative  and  becomes  —  \wl  when 

x  —  L  The  equation  V=.\wl—  wx 

is  indeed  the  equation  of  a  straight  Fig.  12. 

line,  the  origin  being  at  the  left  support,  and  may  be  plotted 

so  that  the   ordinate  at  any  point  will  represent  the  vertical 

shear   for   the    corresponding  section  of  the  beam,  as  shown 

in  Fig.  12. 

Consider  again  a  simple  beam  as  in  Fig.  13  whose  span  is  12 
feet  and  having  three  loads  of  240,  90,  and  120  pounds,  situated 
3,  4,  and  8  feet  respectively  from 
the  left  support,     By  Art.  16  the 
left  reaction   is  found   to  be   280 
and  the  right  reaction  170  pounds. 
Then  for  any  section  between  the 
left    support    and  the    first    load 
the   vertical    shear   is   V  =  -f-  280 
pounds,  for  a  section  between  the 


Fig.  13. 


first  and  second  loads  it  is  V  =•  280  —  240  =  -f-  40  pounds,  for 
a  section  between  the  second  and  third  loads  V  =  280  —  240 
—  90  —  —  50  pounds,,  and  for  a  section  between  the  third  load 
and  the  right  support  V  =  280  —  240  —  90  —  120  =  —  170 
pounds,  which  has  the  same  numerical  value  as  the  right  reac- 
tion. By  laying  off  ordinates  upon  a  horizontal  line  a  graphical 
representation  of  the  distribution  of  vertical  shears  throughout 
the  beam  is  obtained. 

For  any  section  of  a  simple  beam  distant  x  from  the  left 
support,  let  Rt  denote  the  left  reaction,  w  the  weight  of  the 
uniform  load  per  linear  unit,  and  2P^  the  sum  of  all  the  con- 


42  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

centrated  loads  between  the  section  and  that  support.     Then 
the  definition  gives, 


fe 


as  a  general  expression  for  the  vertical  shear  at  that  section. 

A  cantilever  beam  can  be  so  drawn  that  there  is  no  reaction 
at  the  left  end,  and  for  any  section  V  =  —  wx  —  2P, . 

Thus,  in  Fig.  14,  the  vertical 
shear  for  a  section  in  the 
space  a  is  V  =  —  wx,  and  for 
a  section  in  the  space  b  it  is 
V  =  —ivx—P,  and  the  graphi- 
cal representation  is  as  shown 
Fig.  M.  below  the  beam. 

The  vertical  shear  for  any  section  of  a  beam  is  a  measure  of 
the  tendency  to  shearing  along  that  section.  The  above  ex- 
amples show  that  this  is  greatest  near  the  supports.  It  is  rare 
that  beams  actually  fail  in  this  manner,  but  it  is  often  necessary 
to  investigate  the  tendency  to  such  failure. 

Prob.  30.  A  simple  beam  12  feet  long  and  weighing  20 
pounds  per  linear  foot  has  loads  of  600  and  300  pounds  at  2 
and  4  feet  respectively  from  the  left  end.  Find  the  vertical 
shears  at  several  sections  throughout  the  beam,  and  draw  a 
diagram  to  show  their  distribution. 

ART.  1 8.    THE  BENDING  MOMENT. 

The  usual  method  of  failure  of  beams  is  by  cross-breaking  or 
transverse  rupture.  This  is  caused  by  the  external  forces 
producing  rotation  around  some  point  in  the  section  of  failure, 
Thus,  in  Fig.  14,  let  a  be  the  distance  between  the  end  and  the 
load  P,  and  b  be  the  distance  between  P  and  the  wall.  Then 
the  tendency  of  Pto  cause  rotation  around  a  point  in  the  section 
at  the  wall  is  measured  by  its  moment  Pb ;  if,  however,  the 


ART.    1  8. 


THE   BENDING   MOMENT. 


43 


load  were  at  the  end  its  tendency  to  produce  rotation  around 
the  same  point  would  be  measured  by  the  moment  P(a  -(-  b). 

'  Bending  moment '  is  the  name  given  to  the  algebraic  sum 
of  the  moments  of  the  external  forces  on  the  left  of  the  sec- 
tion with  reference  to  a  point  in  that  section.  Let  it  be  de- 
noted by  M.  Then,  for  a  cantilever  or  simple  beam, 

M=  moment  of  reaction  minus  sum  of  moments  of  loads. 
Here  the  moment  of  upward  forces  is  taken  as  positive  and 
that  of  downward  forces  as  negative.     M  may  hence  be  positive 
or  negative  according  as  the  first  or  second  term  is  the  greater. 

For  a  simple  beam  of  length  /,  uniformly  loaded,  each  reaction 
is  fywl.  For  any  section  distant  x  from  the  left  support  the 
moment  is  M  =  \wl .  x  —  wx .  ^x, 
x  being  the  lever  arm  of  the  re- 
action \wl,  and  \x  the  lever  arm 
of  the  load  wx.  Here  M  =  o 
when  x  =  o  and  also  when  x  =  /, 
and  M  is  a  maximum  when  x  =  •£/.  Fie-  »s- 

The  equation,  in  short,  is  that  of  a  parabola  whose  maximum 
ordinate  is  \wl*  and  whose  graphical  representation  is  as  given 
in  Fig.  15,  each  ordinate  showing  the  value  of  M  for  the  cor- 
responding value  of  the  abscissa  x. 

Consider  next  a  simple  beam  loaded  with  only  three  weights 
P,,Py,  and  P3 .  Here  for  any  section  between  the  left  support  and 
the  first  load  M  •=  Rx,  and  for  any  \PL  ip2 

section  between  the  first  and  second    y     "1      I 
loads  M  =  Rx-P,  (x-a).    Each  of 

these  expressions  is  the  equation  of  j  Jllllllllllllilllillli^  I 
a  straight  line,  x  being  the  abscissa 
and  Mthe  ordinate,  and  the  graphical 
representation  of  bending  moments  is  as  shown  in  Fig.  16.  It 
is  seen  that  for  a  simple  beam  all  the  bending  moments  are 
positive. 


.p 
4 


44  CANTILEVER   BEAMS   AND   SIMPLE  BEAMS.      CH.  III. 

For  a  cantilever  there  is  no  reaction  at  the  left  end  and  all 
the  bending  moments  are  negative,  the  tendency  to  rotation 
thus  being  opposite  in  direction  to  that  in  a  simple  beam.  For 
instance,  for  a  cantilever  beam  uniformly  loaded  and  having  a 
load  at  the  end  the  bending  moment  is  M  =  —Px—^wx\ 
Here  the  variation  of  moments  may  be  represented  by  a  parab- 
ola, M  being  o  at  the  free  end  and  a  maximum  at  the  wall. 

For  any  given  case  the  bending  moment  at  any  section  may 
be  found  by  using  the  definition  given  above.  The  external 
forces  on  the  left  of  the  section  are  taken  merely  for  conven- 
ience, for  those  upon  the  right  have  also  the  same  bending 
moment  with  reference  to  the  section.  The  bending  moment 
in  all  cases  is  a  measure  of  the  tendency  of  the  external  forces 
on  either  side  of  the  section  to  turn  the  beam  around  a  point 
in  that  section. 

The  bending  moment  is  a  compound  quantity  resulting  from 
the  multiplication  of  a  force  by  a  distance.  Usually  the  forces 
are  expressed  in  pounds  and  the  distances  in  feet  or  inches  ; 
then  the  bending  moments  are  pound-feet  or  pound-inches. 
Thus  if  a  load  of  500  pounds  be  at  the  middle  of  a  simple 
beam  of  8  feet  span,  the  bending  moment  under  the  load  is, 

M  =  250  X  4  =  i  ooo  pound-feet  =  12  ooo  pound-inches. 
Again  let  a  simple  beam  of  8  feet  span  be  uniformly  loaded 
with  500  pounds  and  have  a  weight  of  200  pounds  at  the  mid- 
dle.    Then  the  bending  moment  at  the  middle  is, 

M  =350  X  4—250  X  2  =  900  pound-feet. 
Hence  the  tendency  to  rupture  is  less  in  the  second  case  than 
in  the  first. 

Prob.  31.  A  beam  6  feet  long  and  weighing  20  pounds  per 
foot  is  placed  upon  a  single  support  at  its  middle.  Compute  the 
bending  moments  for  sections  distant  I,  2,  3,  4,  and  5  feet  from 
the  left  end,  and  draw  a  curve  to  show  the  distribution  of  mo- 
ments throughout  the  beam. 


ART.  19.    INTERNAL   STRESSES   AND   EXTERNAL   FORCES.          45 

Prob.  32.  A  simple  beam  of  6  feet  span  weighs  20  pounds 
per  linear  foot  and  has  a  load  of  270  pounds  at  2  feet  from  the 
left  end.  Find  the  vertical  shears  for  sections  one  foot  apart 
throughout  the  beam,  and  draw  the  diagram  of  shears.  Find 
the  bending  moments  for  the  same  sections  and  draw  the  dia- 
gram to  represent  them. 

ART.  19.    INTERNAL  STRESSES  AND  EXTERNAL  FORCES. 

The  external  loads  and  reactions  on  a  beam  maintain  their 
equilibrium  by  means  of  internal  stresses  which  are  generated 
in  it.  It  is  required  to  determine  the  relations  between  the  ex- 
ternal forces  and  the  internal  stresses ;  or,  since  the  effect  of 
the  external  forces  upon  any  section  is  represented  by  the  ver- 
tical shear  (Art.  17)  and  by  the  bending  moment  (Art.  18),  the 
problem  is  to  find  the  relation  between  these  quantities  and  the 
internal  stresses  in  that  section. 

Consider  a  beam  of  any  kind  which  is  loaded  in  any  manner. 
Imagine  a  vertical  plane  mn  cutting  the  beam  at  any  cross-sec- 
tion.    In  that  section  there  are  act-  i       | 
ing  unknown  stresses  of  various  in- 
tensities and    directions.     Let  the 
beam  be  imagined  to  be  separated 
into  two  parts  by  the  cutting  plane 
and  let  forces  X,  Y,  Z,  etc.,  equiv- 
alent to   the    internal    stresses,  be 
applied  to  the  section  as  shown  in     '  | 

Fig.  17.     Then  the  equilibrium  of     '    x      >jm' 

each  part  of  the  beam  will  be  undis-          y^i*< 

turbed,  for  each  part  will  be  acted         z-*      :» 

upon  by  a  system  of  forces  in  equi- 
librium.    Hence  the  following  fun-  Fig.  I7. 
damental  principle  is  established. 

The  internal  stresses  in  any  cross-section  of  a  beam  hold  in 
equilibrium  the  external  forces  on  each  side  of  that  section. 


1   1    !" 

I       I     m\:       ^ 

r  i u 


f 


46  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

This  is  the  most  important  principle  in  the  theory  of  flexure. 
It  applies  to  all  beams,  whether  the  cross-section  be  uniform  or 
variable  and  whatever  be  the  number  of  the  spans  or  the  na- 
ture of  the  loading. 

Thus  in  the  above  figure  the  internal  stresses  X,  Y,  Z,  etc., 
hold  in  equilibrium  the  loads  and  reactions  on  the  left  of  the 
section,  and  also  those  on  the  right.    Considering  one  part  only 
a  system  of  forces  in  equilibrium  is  seen,  to  which  the  three 
necessary  and  sufficient  conditions  of  statics  apply,  namely, 
2  of  all  horizontal  components  =  o, 
2  of  all  vertical  components  =  o, 
2  of  moments  of  all  forces  =  o. 

From  these  conditions  can  be  deduced  three  laws  concerning 
the  unknown  stresses  in  any  section.  Whatever  be  the  inten- 
sity and  direction  of  these  stresses,  let  each  be  resolved  into 
its  horizontal  and  vertical  components.  The  horizontal  com- 
ponents will  be  applied  at  different  points  in  the  cross-section, 
FJ  ||  some  acting  in  one  direc- 

* J * — i tion  and  some  in  the  other, 

or  in  other  words,  some  of 
the  horizontal    stresses  are 
tensile  and   some  compres- 
Fi«- l8-  sive  :  by  the  first  condition 

the  algebraic  sum  of  these  is  zero.  The  vertical  components 
will  add  together  and  form  a  resultant  vertical  force  V which, 
by  the  second  condition,  equals  the  algebraic  sum  of  the  exter- 
nal forces  on  the  left  of  the  section.  As  this  internal  force 
Facts  in  contrary  directions  upon  the  two  parts  into  which  the 
beam  is  supposed  to  be  separated,  it  is  of  the  nature  of  a  shear. 
Hence  for  any  section  of  any  beam  the  following  laws  concern- 
ing the  internal  stresses  may  be  stated. 

1st.  The  algebraic  sum  of  the  horizontal  stresses  is  zero  ;  or 
the  sum  of  the  horizontal  tensile  stresses  is  equal  to  the 
sum  of  the  horizontal  compressive  stresses. 


ART.  19.    INTERNAL  STRESSES  AND   EXTERNAL  FORCES.        47 

2nd.  The  algebraic  sum  of  the  vertical  stresses  forms  a  re- 
sultant shear  which  is  equal  to  the  algebraic  sum  of  the 
external  vertical  forces  on  either  side  of  the  section. 

3rd.  The  algebraic  sum  of  the  moments  of  the  internal 
stresses  is  equal  to  the  algebraic  sum  of  the  moments  of 
the  external  forces  on  either  side  of  the  section. 

These  three  theoretical  laws  are  the  foundation  of  the  theory 
of  the  flexure  of  beams.  Their  expression  may  be  abbreviated 
by  introducing  the  following  definitions. 

'  Resisting  shear f  is  the  name  given  to  the  algebraic  sum  of 
the  internal  vertical  stresses  in  any  section,  and  '  vertical  shear ' 
is  the  name  for  the  algebraic  sum  of  the  external  vertical  forces 
on  the  left  of  the  section.  '  Resisting  moment '  is  the  name 
given  to  the  algebraic  sum  of  the  moments  of  the  internal  hori- 
zontal stresses  with  reference  to  a  point  in  the  section,  and 
'bending  moment'  is  the  name  for  the  algebraic  sum  of  the 
moments  of  the  external  forces  on  either  side  of  the  section 
with  reference  to  the  same  point.  Then  the  three  laws  may 
be  thus  expressed  for  any  section  of  any  beam, 

Sum  of  tensile  stresses  =  Sum  of  compressive  stresses. 
Resisting  shear  =  Vertical  shear. 

Resisting  moment         =  Bending  moment. 

The  second  and  third  of  these  equations  furnish  the  funda- 
mental laws  for  investigating  beams.  They  state  the  relations 
between  the  internal  stresses  in  any  section  and  the  external 
forces  on  either  side  of  that  section.  For  the  sake  of  uniform- 
ity the  external  forces  on  the  left  hand  side  of  the  section  will 
generally  be  used,  as  was  done  in  Arts.  17  and  18. 

Prob.  33  A  wooden  beam  12  x  14  inches  and  6  feet  long  is 
sustained  at  one  end  by  a  force  of  280  pounds  acting  at  an 
angle  of  60  degrees  with  the  vertical,  and  at  the  other  end  by  a 
vertical  force  Y  and  a  horizontal  force  X,  Find  the  values  of 
X  and  Y. 


48  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 


ART.  20.    EXPERIMENTAL  AND  THEORETICAL  LAWS. 

From  the  three  necessary  conditions  of  static  equilibrium,  as 
stated  in  Art.  19,  three  important  theoretical  laws  regarding 
internal  stresses  were  deduced.  These  alone,  however,  are  not 
sufficient  for  the  full  investigation  of  the  subject,  but  recourse 
must  be  had  to  experience  and  experiment.  Experience  teaches 
that  when  a  beam  deflects  one  side  becomes  concave  and  the 
other  convex,  and  it  is  reasonable  to  suppose  that  the  hori- 
zontal tensile  stresses  are  on  the  convex  side  and  the  compres- 
sive  stresses  on  the  concave.  By  experiments  on  beams  this  is 
confirmed  and  the  following  laws  deduced. 

(F) — The  horizontal  fibers  on  the  convex  side  are  elongated 
and  those  on  the  concave  are  shortened,  while  near  the 
center  is  a  '  neutral  surface '  which  is  unchanged  in  length. 

(G) — The  amount  of  elongation  or  compression  of  any  fiber 
is  directly  proportional  to  its  distance  from  the  neutral 
surface.  Hence  by  law  (B]  the  horizontal  stresses  are 
also  directly  proportional  to  their  distances  from  the 
neutral  surface,  provided  the  elastic  limit  of  the  material 
be  not  exceeded. 

From  these  laws  there  will  now  be  deduced  the  following  im- 
portant theorem  regarding  the  position  of  the  neutral  surface : 

The  neutral  surface  passes  through  the  centers  of  gravity 
of  the  cross-sections. 

To  prove  this  let  a  be  the  area  of  any  elementary  fiber  and  z  its 
distance  from  the  neutral  surface.  Let  S  be  the  unit-stress 
on  the  fiber  most  remote  from  the  neutral  surface  at  the 
distance  c.  Then  by  law  (G), 

—  =  unit-stress  at  the  distance  unity, 

—  z  —  unit-stress  at  the  distance  z, 


ART.  21.        THE   TWO   FUNDAMENTAL   FORMULAS.  49 

therefore      —  az  =  the  total  stress  on  any  fiber  of  area  a, 
and  2 =  algebraic  sum  of  all  horizontal  stresses. 

But  by  the  first  law  of  Art.  19  this  algebraic  sum  is  zero,  and 
since  5  and  c  are  constants  the  quantity  2az  is  also  zero.  This, 
however,  is  the  condition  which  makes  the  line  of  reference 
pass  through  the  center  of  gravity  as  is  plain  from  the  defini- 
tion of  the  term  '  center  of  gravity.'  Therefore,  the  neutral 
surface  of  beams  passes  through 
the  centers  of  gravity  of  the 
cross-sections. 


____._.._        _  Neutral  Surface 

The  'neutral  axis'  of  a  cross- 


section  is  the  line  in  which  the 

neutral    surface    intersects   the  Flg'  I9' 

plane  of  the  cross-section.     On  the  left  of  Fig.  19  is  shown  the 

neutral  axis  of  a  cross-section  and  on  the  right  a  trace  of  the 

neutral  surface. 

Prob.  34.  A  beam  3  inches  wide  and  6  inches  deep  is  loaded 
so  that  the  unit-stress  at  the  remotest  fiber  of  a.  certain  cross- 
section  is  600  pounds  per  square  inch.  Find  the  sum  of  all  the 
tensile  stresses  on  the  cross-section. 

Prob.  35.  A  wooden  beam  6  x  12  inches  and  five  feet  long  is 
supported  at  one  end  and  kept  level  by  two  horizontal  forces 
X  and  Z  acting  at  the  other  end  in  the  median  line  of  the  cross- 
section,  the  former  at  2  inches  from  the  top  and  the  latter  at  2 
inches  from  the  base.  Find  the  values  of  X  and  Z. 


ART.  21.    THE  Two  FUNDAMENTAL  FORMULAS. 

Consider  again  any  beam  loaded  in  any  manner  and  cut  at 
any  section  by  a  vertical  plane.  The  internal  stresses  in  that 
section  hold  in  equilibrium  the  external  forces  on  the  left  of 


5O  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.       CH.  III. 

the  section,  and  as  shown  in  Art.  19,  the  following  fundamental 
laws  obtain, 

Resisting  shear       =  Vertical  shear, 
Resisting  moment  =  Bending  moment. 

The  principles  established  in  the  preceding  pages  can  now  be 
applied  to  the  algebraic  expression  of  these  four  quantities. 

The  resisting  shear  is  the  algebraic  sum  of  all  the  vertical 
components  of  the  internal  stresses  at  any  section  of  the  beam. 
If  A  be  the  area  of  that  section  and  Ss  the  shearing  unit-stress, 
regarded  as  uniform  over  the  area,  then  from  formula  (i), 

Resisting  shear  =  ASS. 

The  vertical  shear  for  the  same  section  of  the  beam  being  V 
(Art.  17),  the  first  of  the  above  fundamental  laws  becomes, 

(3)  AS.  =  V, 

which  is  the  first  fundamental  formula  for  the  discussion  of 
beams. 

The  resisting  moment  is  the  algebraic  sum  of  the  moments 
of  the  internal  horizontal  stresses  at  any  section  with  reference 
to  a  point  in  that  section.  To  find  an  expression  for  its  value 
let  S  be  the  horizontal  unit-stress,  tensile  or  compressive  as  the 
case  may  be,  upon  the  fiber  most  remote  from  the  neutral  axis 
and  let  c  be  the  shortest  distance  from  that  fiber  to  said  axis. 
Also  let  z  be  the  distance  from  the  neutral  axis  to  any  fiber 
having  the  elementary  area  a.  Then  by  law  (G]  and  Fig.  19, 

—  =  unit-stress  at  a  distance  unity, 

£ 

—  z  =  unit-stress  at  distance  2, 

—  =  total  stress  on  any  fiber  of  area  a, 


ART.  21.        THE  TWO   FUNDAMENTAL   FORMULAS.  51 

and  =  moment  of  this  stress  about  neutral  axis. 

c 

2 =  resisting  moment  of  horizontal  stresses. 

Since  5  and  c  are  constants  this  expression  may  be  written 
-'Saz*.  But  2az*,  being  the  sum  of  the  products  formed  by 

multiplying  each  elementary  area  by  the  square  of  its  distance 
from  the  neutral  axis,  is  the  moment  of  inertia  of  the  cross- 
section  with  reference  to  that  axis  and  may  be  denoted  by  /. 
Therefore, 

Resisting  moment  = 

The  bending  moment  for  the  same  section  of  the  beam  being 
M  (Art.  1 8),  the  second  of  the  above  fundamental  laws  becomes, 

(4)  ^  =  W 

which  is  the  second  fundamental  formula  for  the  discussion  of 
beams. 

Experience  and  experiment  teach  that  simple  beams  of  uni- 
form section  break  near  the  middle  by  the  tearing  or  crushing 
of  the  fibers  and  very  rarely  at  the  supports  by  shearing. 
Hence  it  is  formula  (4)  that  is  mainly  needed  in  the  practical 
investigation  of  beams.  The  following  example  and  problem 
relate  to  formula  (3)  only,  while  formula  (4)  will  receive  detailed 
discussion  in  the  subsequent  articles. 

As  an  example,  consider  a  wrought  iron  I  beam  15  feet  long 
and  weighing  200  pounds  per  yard,  over  which  roll  two  locomo- 
tive wheels  6  feet  apart  and  each  bearing  12  ooo  pounds.  The 
maximum  vertical  shear  at  the  left  support  will  evidently  occur 
when  one  wheel  is  at  the  support  (Art.  16).  The  reaction  will 
then  be  500  -[-  12  ooo  +  TV  X  12  ooo  =  19  700  pounds.  Ac- 
cordingJy  the  greatest  value  of  V  in  the  beam  is  19  700 


52  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

pounds.  As  the  area  of  the  cross-section  is  20  square  inches 
the  average  shearing  unit-stress  by  formula  (3)  is  985  pounds, 
so  that  the  factor  of  safety  is  about  50. 

Prob.  36.  A  wooden  beam  6X9  inches  and  12  feet  in  span 
carries  a  uniform  load  of  20  pounds  per  foot  besides  its  own 
weight  and  also  two  wheels  6  feet  apart,  one  weighing  4  ooo 
pounds  and  the  other  3  ooo  pounds.  Find  the  factor  of  safety 
against  shearing. 

ART.  22.  CENTER  OF  GRAVITY  OF  CROSS-SECTIONS. 
The  fundamental  formula  (4)  contains  c,  the  shortest  dis- 
tance from  the  remotest  part  of  the  cross-section  to  a  hori- 
zontal axis  passing  through  the  center  of  gravity  of  that  cross- 
section.  The  methods  of  finding  c  are  explained  in  books  on 
theoretical  mechanics  and  will  not  here  be  repeated.  Its  val- 
ues for  some  of  the  simplest  cases  are  however  recorded  for 
reference. 

For  a  rectangle  whose  height  is  d,  c  =  %d. 

For  a  circle  whose  diameter  is  d,  c  =  \d. 

For  a  triangle  whose  altitude  is  d,  c  =  \d. 

For  a  square  with  side  d  having  one  diagonal 

vertical,  c  = 

For  a  I  whose  depth  is  d,  c  = 

For  a  _L  whose  depth  is  d,  thickness  of  flange  ty 
width  of  flange  b,  and  thickness  of  web  t', 


t'd  +  t(b  —  t'} 
For  a  trapezoid  whose  depth  is  d,  upper  base  b, 

b  +  2b'  d 

and  lower  base  b  .  c  =  —  •  -  •  — 

3 


The  student  should  be  prepared  to  readily  apply  the  principle 
of  moments  to  the  deduction  of  the  numerical  value  of  c  for 
any  given  cross-section.  In  nearly  all  cases  the  given  area 
may  be  divided  into  rectangles,  triangles,  and  circular  areas, 


ART.  23.     MOMENT  OF  INERTIA  OF  CROSS-SECTIONS.  53 

whose  centers  of  gravity  are  known,  so  that  the  statement  of 
the  equation  for  finding  c  is  very  simple. 

Prob.  37.  Find  the  value  of  c  for  a  rail  headed  beam  whose 
section  is  made  up  of  a  rectangular  flange  f  X  4  inches,  a  rect- 
angular web  2-  X  5  inches,  and  an  elliptical  head  £  inches  deep 
and  \\  inches  wide. 

ART.  23.    MOMENT  OF  INERTIA  OF  CROSS-SECTIONS. 

The  fundamental  formula  (4)  contains  /,  the  moment  of  in- 
ertia of  the  cross-section  of  the  beam  with  reference  to  a  hori- 
zontal axis  passing  through  the  center  of  gravity  of  that  cross- 
section.  Methods  of  determining  /  are  explained  in  works  on 
elementary  mechanics  and  will  not  here  be  repeated,  but  the 
values  of  some  of  the  most  important  cases  are  recorded  for 
reference. 

7      73 

For  a  rectangle  of  base  b  and  depth  d,  I  = 

For  a  circle  of  diameter  d,  I  =  -r-  • 

64 

For  an  ellipse  with  axes  a  and  b,  the  latter 

nab* 

vertical,  •  /=-——• 

64 

For  a  triangle  of  base  b  and  depth  d,  1=  —• 

For  a  square  with  side  d,  having  one  diag- 
onal vertical,  7= — -• 

12 

For  a  I  with  base  b,  depth  d,  thickness  of 
flanges  t  and  thickness  of  web  t', 

bd*  —  (b  —  t'}(d '—  2t)> 

12 

For  a  J_  with  base  d,  depth  d,  thickness 
of   flange   /,  thickness  of  web  t'  and 

bd*-(b-t'}(d-tf 
area  A,          I— - " J,  _  ^c\ 


54  CANTILEVER  BEAMS  AND  SIMPLE  BEAMS.      CH.  III. 

The  value  of  /  for  any  given  section  may  always  be  computed 
by  dividing  the  figure  into  parts  whose  moments  of  inertia  are 
known  and  transferring  these  to  the  neutral  axis  by  means  of 
the  familiar  rule  /,  =  /„-(-  AJf,  where  70  is  the  primitive  value 
for  an  axis  through  the  center  of  gravity,  II  the  value  for  any 
parallel  axis,  A  the  area  of  the  figure  and  h  the  distance  be- 
tween the  two  axes. 

Prob.  38.  Compute  the  least  moment  of  inertia  of  a  trape- 
zoid  whose  altitude  is  3  inches,  upper  base  2  inches,  and  lower 
base  5  inches. 

Prob.  39.  Find  the  moment  of  inertia  of  a  triangle  with  ref- 
erence to  its  base,  and  also  with  reference  to  a  parallel  axis 
passing  through  its  vertex. 

ART.  24.    THE  MAXIMUM  BENDING  MOMENT. 

The  fundamental  equation  (4),  namely  —  =  M,  is,  true  for 

any  section  of  any  beam,  /  being  the  moment  of  inertia  of  that 
section  about  its  neutral  axis,  c  the  vertical  distance  from  that 
axis  to  the  remotest  fiber,  S  the  tensile  or  compressive  unit- 
stress  on  that  fiber,  and  M  the  bending  moment  of  all  the  ex- 
ternal forces  on  one  side  of  the  section.  For  a  beam  of  con- 
stant cross-section  5  varies  directly  as  M,  and  the  greatest  5 
will  be  found  where  M  is  a  maximum.  The  place  where  M 
has  its  maximum  value  may  hence  be  called  the  '  dangerous 
section,'  it  being  the  section  where  the  horizontal  fibers  are 
most  highly  strained. 

For  a  simple  beam  uniformly  loaded  with  w  pounds  per 
linear  unit,  the  dangerous  section  is  evidently  at  the  middle, 

W? 

and,  as  shown  in  Art.  18,  the  maximum  Mis  --5-- 

o 

For  a  simple  beam  loaded  with  a  single  weight  P  at  the  dis- 
tance/ from  the  left  support,  the  left  reaction  is  R  =  P — y-^, 


\ 


ART.  24.  THE   MAXIMUM   BENDING   MOMENT.  55 

and  the  maximum  moment  is -, — — .     If  Pbe  movable  the 

distance/ will  be  variable,  and  when  the  load  is  at  the  middle 
the  maximum  M  is  \Pl. 

For  a  beam  loaded  with  given  weights,  either  uniform  or 
concentrated,  it  may  be  shown  that  the  dangerous  section  is  at 
the  point  where  the  vertical  shear  passes  through  zero.  To 
prove  this  let  P1  be  any  concentrated  load  on  the  left  of  the 
section  and  /  its  distance  from  the  left  support,  and  w  the  uni- 
form load  per  linear  unit.  Then,  for  any  section  distant  x 
from  the  left  support, 

M~R.x-wx.Z-  2P,(x  -/). 

To  find  the  value  of  x  which  renders  this  a  maximum,  the  first 
derivative  must  be  put  equal  to  zero  ;  thus, 

£^=  *,-«*-;*/>=  a 

dx 

But  R^  —  wx  —  2Pl  is  the  vertical  shear  V  for  the  section  x 
(see  Art.  17).  Therefore  the  maximum  moment  occurs  at  the 
section  where  the  vertical  shear  passes  through  zero. 

To  find  the  dangerous  section  for  any  given  case  the  reac- 
tions are  first  to  be  computed  by  Art.  16,  and  then  the  verti- 
cal shears  are  to  be  investigated  by  Art.  17.  For  a  cantilever, 
however  it  be  loaded,  it  is  seen  that  the  vertical  shear  becomes 
zero  at  the  wall.  For  a  simple  beam  with  concentrated  loads 
the  point  where  the  vertical  shear  passes  through  zero  must 
usually  be  ascertained  by  trial.  Thus,  referring  to  Fig.  9  and 
the  example  in  Art.  16,  the  vertical  shear  just  at  the  left  of 
the  first  load  is  V  =  520  —  3  X  35  =  +  4J5  pounds,  and  just 
at  the  right  of  the  first  load  it  is  V—  520  —  3  X  35  —  300  = 
+  115  pounds.  Again  for  the  second  load  the  vertical  shear 
just  at  the  left  is  V  •=.  520—  5  X35  —  300  =  -f-  45  pounds,  and 
just  at  the  right  \t\sV=  520  —  5  X  35  —  360  =  — -  15  pounds. 
Hence  in  this  case  the  vertical  shear  changes  sign,  or  passes 


56  CANTILEVER  BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

through  zero,  under  the  second  load,  and  accordingly  this  is 
the  position  of  the  dangerous  section. 

When  the  dangerous  section  has  been  found  the  bending 
moment  for  that  section  is  to  be  computed  by  the  definition 
of  Art.  1 8,  and  this  will  be  the  maximum  bending  moment  for 
the  beam.     Thus,  for  the  numerical  example  of  the  last  para- 
graph, the  maximum  bending  moment  is, 
M=  520  X  5  —  175  X  2j  —  300  X  2  =  +  1562.5  pound-feet. 
Again,  let  a  cantilever  beam  8  feet  long  be  loaded  with  40  pounds 
per  linear  foot  and  carry  a  weight  of  1 50  pounds  at  the  free  end  ; 
then  the  maximum  bending  moment  is, 

M  =  —  320  X  4  —  150x8=—  2  480  pound-feet. 
The  bending  moment  for  simple  beams  is  seen  to  be  always 
positive  and  for  cantilever  beams  always  negative.  That  is  to 
say,  in  the  former  case  the  exterior  forces  on  the  left  of  the 
section  cause  compression  in  the  upper  and  tension  in  the  lower 
fibers  of  the  beam,  while  in  the  latter  case  this  is  reversed ;  or 
the  upper  side  of  a  deflected  simple  beam  is  concave  and  the 
upper  side  of  a  deflected  cantilever  beam  is  convex. 

Prob.  40.  A  simple  beam  12  feet  long  carries  a  load  of  150 
pounds  at  5  feet  from  the  left  end  and  a  load  of  150  pounds  at 
5  feet  from  the  right  end.  Find  the  dangerous  section,  and  the 
maximum  bending  moment. 

Prob.  41.  A  simple  beam  12  feet  long  weighs  20  pounds  per 
foot  and  carries  a  load  of  100  pounds  at  4  feet  from  the  left  end 
and  a  load  of  50  pounds  at  7  feet  from  the  left  end.  Find  the 
dangerous  section,  and  the  maximum  bending  moment. 

ART.  25.    THE  INVESTIGATION  OF  BEAMS. 

The  investigation  of  a  beam  consists  in  deducing  the  greatest 
horizontal  unit-stress  5  in  the  beam  from  the  fundamental  for- 
mula (4).  This  may  be  written, 


ART.  25.  THE    INVESTIGATION  OF  BEAMS.  57 

First,  from  the  given  dimensions  find,  by  Art.  22,  the  value  of  c 
and  by  Art.  23  the  value  of  /.  Then  by  Art.  24  determine  the 
value  of  maximum  M.  From  (4)  the  value  of  5  is  now  known. 
Usually  c  and  /  are  taken  in  inches,  and  M  in  pound-inches ; 
then  the  value  of  5  will  be  in  pounds  per  square  inch. 

The  value  of  5"  will  be  tension  or  compression  according  as 
the  remotest  fiber  lies  on  the  concave  or  convex  side  of  the  beam. 
If  S'  be  the  unit-stress  on  the  opposite  side  of  the  beam  and 
c'  the  distance  from  the  neutral  axis,  then  from  law  (G), 

S      S'  cef 

-  =  — r  and  5    =  S—  . 

c        c  c 

If  6"  be  tension,  S'  will  be  compression,  and  vice  versa.  Some- 
times it  is  necessary  to  compute  S'  as  well  as  5  in  order  to 
thoroughly  investigate  the  stability  of  the  beam.  By  comparing 
the  values  of  5  and  S'  with  the  proper  working  unit-stresses 
for  the  given  materials  (Art.  8),  the  degree  of  security  of  the 
beam  may  be  inferred. 

As  an  example  consider  a  wrought  iron  I  beam  whose  depth 
is  12  inches,  width  of  flange  4.5  inches,  thickness  of  flange  I  inch 
and  thickness  of  web  0.78  inches.  It  is  supported  at  its  ends 
forming  a  span  of  12  feet,  and  carries  two  loads  each  weighing 
10000  pounds,  one  being  at  the  middle  and  the  other  at  one 
foot  from  the  right  end. 

By  Art.  i,  w  =  $6  pounds  per  linear  foot. 

By  Art.  16,  R  =  6169  pounds. 

By  Art.  22,  c  =  6  inches. 

By  Art.  23,  /=  338  inches4. 

By  Art.  24,  x  =.  6  feet  for  dangerous  section. 

By  Art.  24,  max.  M  =  36006  x  12  pound-inches. 

Then  from  formula  (4)  the  unit-stress  at  the  dangerous  section  is, 

~      36000  x  12  x  6 

5  = =  7  700  pounds  per  square  inch. 

33° 


58  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 

This  is  the  compressive  unit-stress  on  the  upper  fiber  and  also 
the  tensile  unit-stress  on  the  lower  fiber,  and  being  only  about 
one-third  of  the  elastic  limit  for  wrought  iron  and  about  one- 
seventh  of  the  ultimate  strength  it  appears  that  the  beam  is 
entirely  safe  for  steady  loads  (Art.  8).  It  will  usually  be  best 
in  solving  problems  to  insert  all  the  numerical  values  at  first  in 
the  formula  and  thus  obtain  the  benefit  of  cancellation. 

A  short  beam  heavily  loaded  should  also  be  investigated  for 
the  shearing  stress  at  the  supports  in  the  manner  mentioned  in 
Art.  21,  but  in  ordinary  cases  there  is  little  danger  from  this 
cause.  Thus  for  the  above  example  the  maximum  vertical 
shear  occurs  at  the  right  end  and  is  14  500  pounds ;  as  the 
area  of  the  cross-section  is  16.8  square  inches,  the  mean  shear- 
ing unit-stress  at  the  right  end  is  from  (3), 

.S  =  --^ —  =  863  pounds  per  square  inch, 
IO.8 

so  that  the  factor  of  safety  against  shearing  is  nearly  60. 

Prob.  42.  A  piece  of  scantling  2  inches  square  and  10  feet 
long  is  hung  horizontally  by  a  rope  at  each  end  and  three 
painters  stand  upon  it.  Is  it  safe  ? 

Prob.  43.  A  wrought  iron  bar  one  inch  in  diameter,  and  two 
feet  long  is  supported  at  its  middle  and  a  load  of  500  pounds 
hung  upon  each  end  of  it.  Find  its  factor  of  safety. 

ART.  26.    SAFE  LOADS  FOR  BEAMS. 

The  proper  load  for  a  beam  should  not  make  the  value  of  5 
at  the  dangerous  section  greater  than  the  allowable  unit-stress. 
This  allowable  unit-stress  or  working  strength  is  to  be  assumed 
according  to  the  circumstances  of  the  case  by  first  selecting  a 
suitable  factor  of  safety  from  Art.  8  and  dividing  the  ultimate 
strength  of  the  material  by  it,  the  least  ultimate  strength  whether 
tensile  or  compressive  being  taken.  For  any  given  beam  the 
quantities  /  and  c  are  known.  Then,  by  the  general  formula  (4), 


ART.  27.  DESIGNING  OF  BEAMS.  59 

the  bending  moment  M  may  be  expressed  in  terms  of  the  un- 
known loads  on  the  beam,  and  thus  those  loads  be  found.  The 
sign  of  the  bending  moment  should  not  be  used  in  (4),  since 
that  sign  merely  denotes  whether  the  upper  fiber  of  the  beam 
is  in  tension  or  compression,  or  indicates  the  direction  in  which 
the  external  forces  tend  to  bend  it. 

As  an  example,  consider  a  cantilever  beam  whose  length  is 
6  feet,  breadth  2  inches,  depth  3  inches  and  which  is  loaded 
uniformly  with  w  pounds  per  linear  foot.  It  is  required  to  find 
the  value  of  w  so  that  .S  may  be  800  pounds  per  square  inch. 
Here  c  =  i^  inches,  /=  ^,  and  M  =  36  x  6zv.  Then  from 
formula  (4), 

800  X  54 

2iozy—  — ; ,          whence        w  =  n  pounds. 

i£x  12  ' 

Since  a  wooden  beam  2X3  inches  weighs  about  2  pounds  per 
linear  foot,  the  safe  load  in  this  case  will  be  about  9  pounds 
per  foot. 

Prob.  44.  A  wooden  beam  8x9  inches  and  of  14  feet  span 
carries  a  load,  including  its  own  weight,  of  w  pounds  per  linear 
foot.  Find  the  value  of  w  for  a  factor  of  safety  of  10. 

Prob.  45.  A  steel  railroad  rail  of  2  feet  span  carries  a  load 
P  at  the  middle.  If  its  weight  per  yard  is  56  pounds,  /=  12.9 
inches4  and  c=  2.16  inches,  find  P  so  that  the  greatest  horizon- 
tal unit-stress  at  the  dangerous  section  shall  be  6  ooo  pounds 
per  square  inch. 

ART.  27.    DESIGNING  OF  BEAMS. 

When  a  beam  is  to  be  designed  the  loads  to  which  it  is  to  be 
subjected  are  known,  as  also  is  its  length.  Thus  the  maximum 
bending  moment  may  be  found.  The  allowable  working  strength 
6"  is  assumed  in  accordance  with  engineering  practice.  Then 
formula  (4)  may  be  written, 

/      M 


60  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

and  the  numerical  value  of  the  second  member  be  found.  The 
dimensions  to  be  chosen  for  the  beam  must  give  a  value  of- 

equal  to  this  numerical  value,  and  these  in  general  are  deter- 
mined tentatively,  certain  proportions  being  first  assumed.  The 
selection  of  the  proper  proportions  and  shapes  of  beams  for 
different  cases  requires  much  judgment  and  experience.  But 
whatever  forms  be  selected  they  must  in  each  case  be  such  as 
to  satisfy  the  above  equation. 

For  instance,  a  wrought  iron  beam  of  4  feet  span  is  required 
to  carry  a  rolling  load  of  500  pounds.  Here,  by  Art.  24,  the 
value  of  maximum  M  due  to  the  load  of  500  pounds  is  6  ooo 
pound-inches.  From  Art.  8  the  value  of  6"  for  a  variable  load 
is  about  10000  pounds  per  square  inch.  Then, 

/       6000 


10000 


=  0.6  inches3. 


An  infinite  number  of  cross-sections  may  be  selected  with  this 

value  of  -.     If  the  beam  is  to  be  round  and  of  diameter  d.  it 
c 

is  known  that  c  =  \d  and  f=  -?—.     Hence, 

04 

nd* 

—  =0.6,        whence  d=  1.83  inches. 

If  the  cross-section  is  to  be  rectangular,  the  dimensions  1x2 
inches  would  give  the  value  of  —  as  £  which  would  be  a  little 

too  large,  but  it  would  be  well  to  use  it  because  the  weight  of 
the  beam  itself  has  not  been  considered  in  the  discussion.  If 
thought  necessary  these  dimensions  may  now  be  investigated 
by  Art.  25  in  order  to  determine  how  closely  the  actual  unit- 
stress  agrees  with  the  value  assumed.  Thus  the  rectangular 
section  I  x  2  inches  weighs  6-f  pounds  per  foot ;  the  maximum 


ART.  28.  THE   MODULUS   OF  RUPTURE.  6l 

bending  moment  is  then  6  1 60  pound-inches,  and  the  unit-stress 
is  found  to  be  9  240  pounds  per  square  inch. 

Prob.  46.  Design  a  hollow  circular  wrought  iron  beam  for  a 
span  of  12  feet  to  carry  a  load  of  320  pounds  per  linear  foot. 

Prob.  47.  A  rectangular  wooden  beam  of  14  feet  span  carries 
a  load  of  I  ooo  pounds  at  its  middle.  If  its  width  is  4  inches 
find  its  depth  for  a  factor  of  safety  of  10. 

ART.  28.    THE  MODULUS  OF  RUPTURE. 

The  fundamental  formula  (4)  is  only  true  for  stresses  within 
the  elastic  limit,  since  beyond  that  limit  the  law  (G)  does  not 
hold,  and  the  horizontal  unit-stresses 
are  no  longer  proportional  to  their 
distances  from  the  neutral  axis,  but 
increase  in  a  more  rapid  ratio.  The 
sketch  shows  a  case  where  the  fiber 


stresses   above  m  and    below  n  have  Flg-  20- 

surpassed  the  elastic  limit.     It  is  however  very  customary  in 
practical  computations  to  apply  (4)  to  the  rupture  of  beams. 

The  '  modulus  of  rupture '  is  the  value  of  5  deduced  from 
formula  (4)  when  the  beam  is  loaded  up  to  the  breaking  point. 
It  is  always  found  by  experiment  that  the  modulus  of  rupture 
does  not  agree  with  either  the  ultimate  tensile  or  compressive 
strength  of  the  material  but  is  intermediate  between  them.  If 
formula  (4)  were  valid  beyond  the  elastic  limit,  the  value  of  .S 
for  rupture  would  agree  with  the  least  ultimate  strength,  with 
tension  in  the  case  of  cast  iron  and  with  compression  in  the  case 
of  timber.  The  modulus  of  rupture  is  denoted  by  Sr . 

The  average  values  of  the  modulus  of  rupture  are  given  in 
the  following  table,  which  also  contains  the  average  ultimate 
tensile  and  compressive  strengths,  previously  stated  in  Arts.  5 
and  6,  all  in  pounds  p.er  square  inch. 


62 


CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 


Material. 

Tensile 
Strength,  St  . 

Modulus  of 
Rupture,  Sr  . 

Compressive 
Strength,  Sc  . 

Timber 

IOOOO 

9OOO 

8000 

Brick 

800 

2  5OO 

Stone 

2000 

6OOO 

Cast  Iron 

20000 

35000 

90000 

Wrought  Iron 

55000 

55000 

55000 

Steel 

IOOOOO 

I2OOOO 

150000 

By  the  use  of  the  experimental  values  of  the  modulus  of  rup- 
ture it  is  easy  with  the  help  of  formula  (4)  to  determine  what 
load  will  cause  the  rupture  of  a  given  beam,  or  what  must  be 
its  length  or  size  in  order  that  it  may  rupture  under  assigned 
loads.  The  formula  when  used  in  this  manner  is  entirely  em- 
pirical and  has  no  rational  basis. 

Prob.  48.  What  must  be  the  size  of  a  square  wooden  beam  of 
8  feet  span  in  order  to  break  under  its  own  weight  ? 

Prob.  49.  A  cast  iron  cantilever  beam  2  inches  square  and  6 
feet  long  carries  a  load  P  at  the  end.  Find  the  value  of  P  to 
cause  rupture. 

ART.  29.    COMPARATIVE  STRENGTHS. 

The  strength  of  a  beam  is  measured  by  the  load  that  it  can 
carry.  Let  it  be  required  to  determine  the  relative  strength  of 
the  four  following  cases, 

ist,  A  cantilever  loaded  at  the  end  with  W, 

2nd,  A  cantilever  uniformly  loaded  with  W, 

3rd,  A  simple  beam  loaded  at  the  middle  with  W, 

4th,  A  simple  beam  loaded  uniformly  with  W. 

Let  /  be  the  length  in  each  case.     Then,  from  Art.  24  and  for< 
mula  (4), 

For  ist,     M=Wl    and  hence      W=~. 


ART.  29.  COMPARATIVE  STRENGTHS.  63 

For  2nd,       M  =  —         and  hence          W=2-^-. 

For  3rd,        M  —  —          and  hence          W=  4-7-. 
4  ic 

For  4th,        M=-Q-         and  hence          W  =  8-v-. 

o  *£ 

Therefore  the  comparative  strengths  of  the  four  cases  are  as 
the  numbers  I,  2,  4,  8.  That  is,  if  four  such  beams  be  of  equal 
size  and  length  and  of  the  same  material,  the  2nd  is  twice  as 
strong  as  the  1st,  the  3rd  fourtimes  as  strong,  and  the  4th  eight 
times  as  strong.  From  these  equations  also  result  the  follow- 
ing important  laws. 

The  strength  of  a  beam  varies  directly  as  S,  directly  as 
/,  inversely  as  c,  and  inversely  as  the  length  /. 

A  load  uniformly  distributed  produces  only  one-half  as 

much  stress  as  the  same  load  when  concentrated. 
These  apply  to  all  cantilever  and  simple  beams  whatever  be 
the  shape  of  the  cross-section. 

When  the  cross-section  is  rectangular,  let  b  be  the  breadth 
and  ^the  depth,  then  (Art.  23)  the  above  equations  become, 

SbcT 

W=n~6T> 

where  n  is  either  I,  2, 4,  or  8,  as  the  case  may  be.     Therefore, 
The  strength  of  a  rectangular  beam  varies  directly  as  the 
breadth  and  directly  as  the  square  of  the  depth. 

The  reason  why  rectangular  beams  are  put  with  the  greatest 
dimensions  vertical  is  now  apparent. 

To  find  the  strongest  rectangular  beam  that  can  be  cut  from 
a  circular  log  of  given  diameter  D,  it  is  necessary  to  make  bd* 
a  maximum.  Or  the  value  of  b  is  to  be  found  which  makes 
b  (D*  —  £')  a  maximum.  By  placing  the  first  derivative  equal 
to  zero  this  value  of  b  is  readily  found.  Thus, 
and  d 


Fig.  21. 


64  CANTILEVER  BEAMS   AND   SIMPLE  BEAMS.      CH.  III. 

Hence  very  nearly,  b :  d : :  5  : 7.  From  this  it  is  evident 
that  the  way  to  lay  off  the  strongest  beam  on 
the  end  of  a  circular  log  is  to  divide  the  diameter 
into  three  equal  parts,  from  the  points  of  divi- 
sion draw  perpendiculars  to  the  circumference, 
and  then  join  the  points  of  intersection  with  the 
ends  of  the  diameter,  as  shown  in  the  figure. 
The  beam  thus  cut  out  is,  of  course,  not  as  strong  as  the  log, 
and  the  ratio  of  the  strength  of  the  beam  to  that  of  the  log  is 

that  of  their  values  of  -,  which  will  be  found  to  be  about  0.65. 

Prob.  50.  Compare  the  strength  of  a  rectangular  beam  2 
inches  wide  and  4  inches  deep  with  that  of  a  circular  beam  3 
inches  in  diameter. 

Prob.  51.  Compare  the  strength  of  a  wooden  beam  4  X  6 
inches  and  10  feet  span  with  that  of  a  wrought  iron  beam  1X2 
inches  and  7  feet  span. 


ART.  30.    WROUGHT  IRON  I  BEAMS. 

Wrought  iron  I  beams  are  rolled  at  present  in  about  thirteen 
different  depths  or  sizes ;  of  each  there  is  a  light  and  a  heavy 
-j-46^.^— ..»   ^gh^  ancj  Weights  intermediate   in   value 
may  also  be  obtained.     They  are  extensively 
used  in  engineering  and  architecture.     The 
following   table  gives   mean   sizes,  weights, 
and   moments  of    inertia    of    wrought   iron 
beams  most  commonly  found  in  the  market. 
The  sizes  of  different    manufacturers  agree 
}  IL^       as  to  depth,  but  vary  slightly  with  regard  to 
I  f~~~^       ^^^  proportions  of  cross-section,  weights  per  foot, 
Fig.  33.  an(j  moments  of  inertia.     Fig.  22  shows  the 

proportions  of  the  light  and  heavy  6  inch  beams.     The  cross- 
section  of  any  beam  in  the  table  is  obtained  from  its  weight  per 


ART.  30. 


WROUGHT   IRON   I   BEAMS. 


foot  by  multiplying  by  3  and  dividing  by  10,  in  accordance  with 
the  rule  of  Art.  i. 

The  moments  of  inertia  in  the  fourth  column  of  the  table  are 
taken  about  an  axis  perpendicular  to  the  web  at  the  center, 
this  being  the  neutral  axis  of  the  cross-section  when  used  as  a 
beam.  The  values  of  /'  are  with  reference  to  an  axis  coincid- 
ing with  the  center  line  of  the  web  and  are  for  use  in  Chapter  V. 


Size. 
Depth. 

Inches. 

Width  of 
Flange. 

Inches. 

Weight 
per  foot. 

Pounds. 

/. 
Inches*. 

/ 
Inches'. 

/'. 
Inches*. 

Heavy   15 

5.8l 

80 

750 

IOO. 

29.9 

Light     15 

5-55 

67 

677 

90-3 

25.4 

H           15 

5-33 

65 

614 

81.9 

20.  o 

L            15 

5-03 

50 

530 

70.6 

16.3 

H              12 

5-09 

60 

340 

56.7 

15.5 

L               12 

4.64 

42 

275 

45-9 

II.  O 

H           ioi 

4.92 

45 

201 

38.3 

10.7 

L           ioi 

4-54 

3ii 

165 

3i-4 

8.01 

H           10 

4-77 

45 

I87 

37-5 

n-3 

L           10 

4-32 

30 

150 

30.0 

7-94 

H            9 

4-33 

33 

H7 

26.0 

7.14 

L             9 

4.01 

23i 

97-5 

21.7 

5.48 

H            8 

4.29 

35 

90.4 

22.6 

6.96 

L             8 

3-8i 

22 

69.6 

17-5 

4-57 

H            7 

3-91 

25 

54-3 

15-5 

4.87 

L             7 

3-6i 

18 

45-8 

I3-I 

3-72 

H            6 

3-46 

18 

28.4 

9.48 

2-51 

L             6 

3-24 

I3i 

24.5 

8.16 

2.00 

H            5 

2.91 

13 

14.2 

5.69 

1-34 

L             5 

2-73 

10 

12.3 

4-94 

i.  08 

H            4 

2.63 

10 

6.99 

3-50 

0.87 

L             4 

2.48 

8 

6.19 

3-10 

0.71 

H            3 

2.52 

9 

3-54 

2.36 

0.84 

L             3 

2.32 

7 

3-09 

2.06 

0-55 

66  CANTILEVER   BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

In  investigating  the  strength  of  a  given  I  beam  the  value  of  - 

is  taken  from  the  table  and  S  is  computed  from  formula  (4). 
In  designing  an  I  beam  for  a  given  span  and  loads  the  value  of 

-  is  found  by.  (4)  from  the  data  and  then  from  the  table  that  I 

is  selected  which  has  the  nearest  or  next  highest  corresponding 
value.     Intermediate  weights  between  those  given  in  the  table 

can  also  usually  be  obtained ;  thus,  if  the  computed  value  of  - 

should  be  34.0  a  lo-inch  beam  weighing  about  38  pounds  per 
foot  might  be  chosen. 

For  example,  let  it  be  required  to  determine  which  I  should 
be  selected  for  a  floor  loaded  with  150  pounds  per  square  foot, 
the  beams  to  be  of  20  feet  span  and  spaced  12  feet  apart  be- 
tween centers,  and  the  maximum  unit-stress  S  to  be  12000 
pounds  per  square  inch.  Here  the  uniform  load  on  the  beam 
is  12  X  20  X  150  =  36000  pounds  =  W.  From  formula  (4), 

I _M  __36ooo  X  20  X  12  _ 
~c~~'S   ~~         8  X  12000 

and  hence  from  the  table,  the  light   1 5  inch  I  should  be  se- 
lected. 

Steel  I  beams  and  other  shapes  are  now  beginning  to  be  used, 
and  will  undoubtedly  be  very  common  in  a  few  years. 

Prob.  52.  A  heavy  15  inch  I  beam  of  12  feet  span  sustains  a 
uniformly  distributed  load  of  41  net  tons.  Find  its  factor  of 
safety.  Also  the  factor  of  safety  for  a  24  feet  span  under  the 
same  load. 

Prob.  53.  A  floor,  which  is  to  sustain  a  uniform  load  of  175 
pounds  per  square  foot,  is  to  be  supported  by  heavy  10  inch  I 
beams  of  15  feet  span.  Find  their  proper  distance  apart  from 
center  to  center  so  that  the  maximum  fiber  stress  may  be 
12000  pounds  per  square  inch. 


ART.  31. 


WROUGHT  IRON   DECK  BEAMS. 


67 


Q 


ART.  31.    WROUGHT  IRON  DECK  BEAMS. 

Deck  beams  are  used  in  the  construction  of  buildings,  and 
are  of  a  section  such  as  shown  in  Fig.  23.  The  heads  are 
formed  with  arcs  of  circles  but  may  be 
taken  as  elliptical  in  computing  the  values 
of  c  and  /.  The  following  table  gives  di- 
mensions of  a  few  wrought  iron  sections 
found  in  the  market. 

By  means  of  formula  (4)  a  given  deck 
beam  may  be  investigated  or  safe  loads  be 
determined  for  it,  or  one  may  be  selected 
for  a  given  load  and  span.  Sometimes  T 
irons  are  used  instead  of  deck  beams ;  the 
values  of  c  and  /  for  these  are  given  in  the  Flg>  23- 

handbooks  issued  by  the  manufacturers,  or  they  may  be  com- 
puted with  an  accuracy  usually  sufficient  by  regarding  the 
web  and  flange  as  rectangular  (Arts.  22  and  23). 


Size.  Depth. 
Inches. 

Width  of 
Flange. 

Inches. 

Thickness 
of  Web. 

Inches.     . 

Weight 
per  ft. 

Pounds. 

Inches. 

/. 

Inches*. 

/ 
c 
Inches*. 

Heavy  9 

3-97 

0.625 

30 

4-59 

91.9 

2O.  O 

Light     9 

3-75 

0.406 

231 

4.60 

78.6 

I7-I 

H            8 

4.00 

0.750 

28 

4-49 

63.3 

I4-I 

L            8 

3-75 

0.500 

2I| 

4-58 

52.1 

ii.  6 

H           7 

3-75 

0.625 

23 

3-98 

43-0 

10.8 

L            7 

3-50 

0-375 

17 

4.00 

34-4 

8.6 

Prob.  54.  A  heavy  7  inch  deck  beam  is  loaded  uniformly 
with  50  ooo  pounds.  Find  its  factor  of  safety  for  a  span  of 
22  feet.  Also  for  a  span  of  1 1  feet. 

Prob.  55.  What  uniform  load  should  be  placed  upon  a  heavy 
7  inch  deck  beam  of  22  feet  span  so  that  the  greatest  unit-stress 
at  the  dangerous  section  may  be  1 2  ooo  pounds  per  square  inch  ? 


68  CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

ART.  32.    CAST  IRON  BEAMS. 

Wrought  iron  beams  are  usually  made  with  equal  flanges 
since  the  resistance  of  wrought  iron  is  about  the  same  for  both 
tension  and  compression.  For  cast  iron,  however,  the  flange 
under  tension  should  be  larger  than  that  under  compression, 
since  the  tensile  resistance  of  the  material  is  much  less  than  its 
compressive  resistance.  Let  S'  be  the  unit-stress  on  the  re- 
motest fiber  on  the  tensile  side  and  5  that  on  the  compressive 
side,  at  the  distances  c'  and  c  respectively  from  the  neutral  axis. 
Then,  from  law  (G), 


Now  if  the  working  values  of  5  and  S'  can  be  selected  the 
ratio  of  c  to  c'  is  known  and  a  cross-section  can  be  designed, 
but  it  is  difficult  to  assign  these  proper  values  on  account  of 
our  lack  of  knowledge  regarding  the  elastic  limits  of  cast  iron. 

According  to   HODGKINSON'S  investigations   the   following 
are  dimensions  for  a  cast  iron  beam  of  equal  ultimate  strength. 

Thickness  of  web  =         t, 

Depth  of  beam  =  13.5^, 

Width  of  tensile  flange  =     I2t, 

Thickness  of  tensile  flange  =       2t, 

Width  of  compressive  flange  =       5/, 
Thickness  of  compressive  flange     =     i%t, 

Value  of  c  =       9/, 

Value  of  /  =  923**. 

Here  the  unit-stress  in  the  tensile  flange  is  one-half  that  in  the 
compressive  flange.  Although  these  proportions  may  be  such 
as  to  allow  the  simultaneous  rupture  of  the  flanges,  yet  it  does 
not  necessarily  follow  that  they  are  the  best  proportions  for 
ordinary  working  stresses,  since  the  factors  of  safety  in  the 
flanges  as  computed  by  the  use  of  formula  (4)  would  be  quite 
different.  The  proper  relative  proportions  of  the  flanges  of 


ART.  32.  CAST  IRON  BEAMS.  69 

cast  iron  beams  for  safe  working  stresses  have  never  been 
definitely  established,  and  on  account  of  the  extensive  use  of 
wrought  iron  the  question  is  not  now  so  important  as  formerly. 

As  an  illustration  of  the  application  of  formula  (4)  let  it  be 
required  to  determine  the  total  uniform  load  W  for  a  cast  iron 
±  beam  of  14  feet  span,  so  that  the  factor  of  safety  may  be  6, 
the  depth  of  the  beam  being  18  inches,  the  width  of  the  flange 
12  inches,  the  thickness  of  the  stem  I  inch,  and  the  thickness 
of  the  flange  i£  inches.  First,  from  Art.  22  the  value  of  c  is 
found  to  be  12.63  inches,  and  that  of  c'  to  be  5.37  inches. 
From  Art.  23  the  value  of  /  is  computed  to  be  I  031  inches4. 
From  Art.  24  the  maximum  bending  moment  is, 

M  =  — -  =  21  W  pound-inches. 

o 

Now  with  a  factor  of  safety  of  6  the  working  strength  5  on  the 
remotest  fiber  of  the  stem  of  the  dangerous  section  is  to  be 

— ^ —  pounds  per  square  inch.     Hence  from  formula  (4), 

2iW  =  90°°°X  *°31,      whence       W=  58  300  pounds. 
6  X  12.63 

Again  with  a  factor  of  safety  of  6  the  working  strength  S'  on 
the  remotest  fiber  of  the  flange  at  the  dangerous  section  is  to  be 
20000 


6 


pounds  per  square  inch.     Hence  from  the  formula, 


2lW  =  20 OOP  X  103^       whence       W  =  30 400  pounds. 

The  total  uniform  load  on  the  beam  should  hence  not  exceed 
30  400  pounds.  Under  this  load  the  factor  of  safety  on  the 
tensile  side  is  6,  while  on  the  compressive  side  it  is  nearly  12. 

Prob.  56.  A  cast  iron  beam  in  the  form  of  a  channel,  or 
hollow  half  rectangle,  is  often  used  in  buildings.  Suppose  the 
thickness  to  be  uniformly  one  inch,  the  base  8  inches,  the  height 


7O  CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

6  inches  and  the  span  12  feet.     Find  the  values  of  5  and  S'  at 
the  dangerous  section  under  a  uniform  load  of  5  ooo  pounds. 


ART.  33.    GENERAL  EQUATION  OF  THE  ELASTIC  CURVE. 

When  a  beam  bends  under  the  action  of  exterior  forces  the 
curve  assumed  by  its  neutral  surface  is  called  the  elastic  curve. 
It  is  required  to  deduce  a  general  expression  for  its  equation. 

Let  pp  in  the  figure  be  any  normal  section  in  any  beam. 
Let  mn  be  any  short  length  dl,  measured  on  the  neutral  surface, 

and  let  qmq  be  drawn  parallel  to 
the  normal  section  through  n. 
Previous  to  the  bending  the  sec- 
tions //  and  tt  were  parallel ; 
now  they  intersect  at  o  the  cen- 
ter of  curvature.  Previous  to 
the  bending  pt  was  equal  to  dl, 
now  it  has  elongated  or  shortened 
the  amount  pq.  The  distance 
pq  will  be  called  A  and  the  dis- 
tance mp  is  the  quantity  c  (Art.  22).  The  elongation  A  is  pro- 
duced by  the  unit-stress  S,  and  from  (2)  its  value  is, 

A  =  — 

where  E  is  the  coefficient  of  elasticity  of  the  material  of  the 
beam.     From  the  similar  figures  omn  and  mpq, 


Fig.  24. 


om 
mn 


pq 


or 


R_ 
dl 


c 

I' 


where  R  is  the  radius  of  curvature  om. 
above  value  of  A,  it  becomes, 


Inserting  in  this  the 


ART.  33.   GENERAL   EQUATION   OF  THE   ELASTIC   CURVE.       71 

But  the  fundamental  formula  (4)  may  be  written  in  the  form 

S_M 
c~  I' 

and  hence,  by  comparison, 


This  is  the  formula  which  gives  the  relation  between  the  bend- 
ing moment  of  the  exterior  forces  and  the  radius  of  curvature 
at  any  section.  Where  M  is  o  the  radius  R  is  oo  ;  where  M  is 
a  maximum  R  has  its  least  value. 

Now,  in  works  on  the  differential  calculus,  the  following 
value  is  deduced  for  the  radius  of  curvature  of  any  plane  curve 
whose  abscissa  is  x,  ordinate  y,  and  length  /,  namely, 

dr 

~  dx .  d*y 
Hence  the  most  general  equation  of  the  elastic  curve  is, 

dr      =  El 
dx  .  d*y    .  M  ' 

which  applies  to  the  flexure  of  all  bodies  governed  by  the  laws 
of  Arts.  3  and  20. 

In  discussing  a  beam  the  axis  of  x  is  taken  as  horizontal  and 
that  of  y  as  vertical.  Experience  teaches  us  that  the  length  of 
a  small  part  of  a  bent  beam  does  not  materially  differ  from 
that  of  its  horizontal  projection.  Hence  dl  may  be  placed 
equal  to  dx  for  all  beams,  and  the  above  equation  reduces  to 
the  form, 

<Ty_       M 
dx*      El' 

This  is  the  general  equation  of  the  elastic  curve,  applicable  to 
all  beams  whatever  be  their  shapes,  loads  or  number  of  spans. 
M  is  the  bending  moment  of  the  external  forces  for  any  sec- 


72  CANTILEVER   BEAMS   AND   SIMPLE   BEAMS.       CH.  III. 

tion  whose  abscissa  is  x,  and  whose  moment  of  inertia  with 
respect  to  the  neutral  axis  is  /.  Unless  otherwise  stated  /  will 
be  regarded  as  constant,  that  is,  the  cross-section  of  the  beam 
is  constant  throughout  its  length. 

To  obtain  the  particular  equation  of  the  elastic  curve  for  any 
special  case,  it  is  first  necessary  to  express  M  as  a  function  of 
x  and  then  integrate  the  general  equation  twice.  The  ordinate 
y  will  then  be  known  for  any  value  of  x.  It  should,  however, 
be  borne  in  mind  that  formula  (5),  like  formula  (4),  is  only  true 
when  the  unit-stress  5  is  less  than  the  elastic  limit  of  the 
material. 

Prob.  57.  A  wooden  beam  £  inch  wide,  f  inch  deep,  and  3 
feet  span  carries  a  load  of  14  pounds  at  the  middle.  Find  the 
radius  of  curvature  for  the  middle,  quarter  points,  and  ends. 

ART.  34.    DEFLECTION  OF  CANTILEVER  BEAMS. 

Case  I.  A  load  at  the  free  end. — Take  the  origin  of  co-ordi- 
nates at  the  free  end,  and  as 
in  Fig.  25,  let  m  be  any  point 
of    the    elastic    curve    whose 
Fjg-  2s-  abscissa  is  x  and  ordinate  y. 

For  this  point  the  bending  moment  M  is  —  Px  and  the  general 
formula  (5)  becomes, 


By  integration  the  first  derivative  is  found  to  be 
dy  Px* 

EITx  =  -  —  +  c 

dv 
But  -j-  is  the  tangent  of  the  angle  which  the  tangent  to  the 

elastic  curve  at  m  makes  with  the  axis  of  x,  and  as  the  beam  is 


ART.  34.    DEFLECTION  OF  CANTILEVER  BEAMS.         73, 

dy 
fixed  at  the  wall  the  value  of  -j-  is  o  when  x  equals  /.     Hence 

C  =  $PI\  and  the  first  differential  equation  is, 


dx~ 

The  second  integration  now  gives, 


But  y  =  o,  when  x  =  o.     Hence  C'  =  o,  and 
6EIy  =  P(zl*x  -  x*), 

which  is  the  equation  of  the  elastic  curve  for  a  cantilever  of 
length  /with  a  load  P  at  the  free  end.  If  x  —  I  the  value  of 
y  will  be  the  maximum  deflection,  which  may  be  represented 
by  A.  Then, 


and  for  any  point  of  the  beam  the  deflection  is  A  —  y. 

Case  II.  A  uniform  load.  —  Let  the  origin  be  taken  at  the 
free  end  as  before,  and   x  and  y 
be  the  co-ordinates  of  any  point 
of  the  elastic  curve.     Let  the  load    ^ 
per  linear  unit  be  w.     Then    for  Fie-  26- 

any  section  M  =  —  fywx*  and  formula  (5)  becomes, 


^ 

"   dx1  2  ' 

Integrate  this,  determine  the  constant  of  integration  by  the 

consideration  that  -=-  =  o  when  x  =  /,  and  then, 
dx 


-=  wr  -wx\ 
dx 


74  CANTILEVER   BEAMS  AND   SIMPLE   BEAMS.       CH.   III. 

Integrate  again,  and  after  determining  the  constant,  the  equa- 
tion of  the  elastic  curve  is, 


which  is  a  biquadratic  parabola.     For  x  =  /,  y  =  A  the  maxi- 
mum deflection,  whose  value  is, 

_  u>r  _  wr 
Wi~Wr 

where  W  is  the  total  uniform  load  on  the  cantilever. 

Case  III.  A  load  at  the  free  end  and  also  a  uniform  load.  — 
Here  it  is  easy  to  show  that  the  maximum  deflection  is 

8 


which  is  the  sum  of  the  deflections  due  to  the  two  loads. 
Hence  it  appears  that,  as  in  cases  of  stress,  each  load  produces 
its  effect  independently  of  the  other. 

In  order  that  the  formulas  for  deflection  may  be  true,  the 
maximum  unit-stress  S  produced  by  all  the  loads  must  not 
exceed  the  elastic  limit  of  the  material. 

Prob.  58.  Compute  the  deflection  of  a  cast  iron  cantilever 
beam,  2X2  inches  and  6  feet  span,  caused  by  a  load  of  100 
pounds  at  the  end. 

Prob.  59.  In  order  to  find  the  coefficient  of  elasticity  of  a 
cast  iron  bar  2  inches  wide,  4  inches  deep,  and  6  feet  long,  it 
was  balanced  upon  a  support  and  a  weight  of  4000  pounds 
hung  at  each  end,  causing  a  deflection  of  0.401  inches.  Com- 
pute the  value  of  E. 

ART.  35.    DEFLECTION  OF  SIMPLE  BEAMS. 

The  deflection  of  a  simple  beam  due  to  a  load  at  the  middle, 

or  to  a  uniform  load,  is  readily  obtained  from  the  expressions 

just  deduced  for  cantilever  beams.     Thus,  for  a  simple  beam 

of  span  /with  a  load  P  at  the  middle,  if  Fig.  27  be  inverted  it 


ART.  35.  DEFLECTION  OF  SIMPLE  BEAMS.  75 

will  be  seen  to  be  equivalent  to  two  cantilever  beams  of  length 
£/  with  a  load  \P  at  each  end.  The  formula  for  the  maximum 
deflection  of  a  cantilever  beam  hence  applies  to  this  figure,  if 

P/3 
/ be  replaced  by  £/ and  P  by  \P,  which  gives  A  =       .-.,  for  the 

deflection  at  the  middle  of  the  simple  beam.  It  will  be  well, 
however,  to  use  the  general  formula  (5)  and  treat  each  case  in- 
dependently. 

Case  I.  A  single  load  P  at  the  middle. — Let  the  origin  be 
taken  at  the  left  support.     For 
any  section    between    the    left 
support     and    the    middle    the 
bending    moment    M    is    \Px.  Fi£-  27- 

Then  the  general  formula  (5)  becomes, 
d*y  _  Px 
dx*        2 

dy 

Integrate  this  and  find  the  constant  by  the  fact  that  -j-  =  o 

when  x  =  % .  Then  integrate  again  and  find  the  constant  by 
the  fact  that  y  =  o  when  x  =  o.  Thus, 


is  the  equation  of  elastic  curve  between  the  left  hand  support 
and  the  load.     For  the  greatest  deflection  make  x  =  £/,  then, 

4=pi' 


Case  II.  A  uniform  load. — Let  w  be  the  load  per  linear  unit, 
then  the  formula  (5)  becomes, 

r-.j-d'^y wlx      wx* 

~dx*  ~  ~2~       ~2~' 

Integrate  this  twice,  find  the  constants  as  in  the  preceding  para- 
graph, and  the  equation  of  the  elastic  curve  is, 

—  w(—  x'  -f-  2/^3  —  l*x}, 


76  CANTILEVER  BEAMS  AND   SIMPLE   BEAMS.      CH.  III. 

from  which  the  maximum  deflection  is  found  to  be, 

$wr       5  wr 

Case  III.  A  load  P  at  any  point. — Here  it  is  necessary  first 
to  consider  that  there  are  two  elastic  curves,  one  on  each  side 

of  the  load,  which  have  distinct 

5^-—- — M J    ^^£    equations,    but    which     have    a 

\R      ^=ss==i ^^  common    tangent   and    ordinate 

under  the  load.     As  in  Fig.  25, 

let  the  load  be  placed  at  a  distance  kl  from  the  left  support, 
k  being  a  number  less  than  unity.     Then  the  left  reaction  is 
R  =  P(i  —  k}.     From  the  general  formula  (5),  with  the  origin 
at  the  left  support,  the  equations  are, 
On  the  left  of  the  load, 

(a)        EI^Z  =  Rx, 

w     £f-£  =  $Ks  +  c1, 


On  the  right  of  the  load, 

(a)'       EI^  =  Rx-P(x-  kl\ 

(bj        EI^  =  \Rx*  -  \Px*  +  Pklx  +  C3 , 

(c)f  Ely  =  \Rx3  -  \Px*  +  ^Pklx1  +  £>  +  C, . 

To  determine  the  constants  consider  in  (c)  that  y  =  o  when 
x  =  o,  and  hence  that  Cy  =  o.  Also  in  (c)r,  y  =  o  when  x  —  /; 
again  since  the  curves  have  a  common  tangent  under  the  load, 
(fy  —  (b}'  when  x  =  kl,  and  since  they  have  a  common  ordinate 
at  that  point  (c)  =  (c)'  when  x  —  kl.  Or, 
o  =  \Rl* 

-f  \ 


ART.  36.   COMPARATIVE  DEFLECTION  AND   STIFFNESS.  // 

From  these  three  equations  the  values  of  £7,  ,  Ca  ,  and  C4  are 
found.  Then  the  equation  of  the  elastic  curve  on  the  left  of 
the  load  is, 

P(i  -  k)x*  -  P(2k  -  3 


To  find  the  maximum  deflection,  the  value  of  x  which  renders 
y  a  maximum  is  to  be  obtained  by  equating  the  first  derivative 
to  zero.  If  k  be  greater  than  £,  this  value  of  x  inserted  in  the 
above  equation  gives  the  maximum  deflection  ;  if  k  be  less  than 
£,  the  maximum  deflection  is  on  the  other  side  of  the  load. 
For  instance,  if  k  =  f  ,  the  equation  of  the  elastic  curve  on  the 
left  of  the  load  is, 


This  is  a  maximum  when  x  =  O-56/,  which  is  the  point  of  great- 
est deflection. 

Prob.  60.  Prove,  when  k  is  greater  than  |-  in  Fig.  28,  that  the 

>" 
maximum  deflection  is  A  = 


Prob.  61.  In  order  to  find  the  coefficient  of  elasticity  of 
Quercus  alba  a  bar  4  centimeters  square  and  one  meter  long 
was  supported  at  the  ends  and  loaded  in  the  middle  with  weights 
of  50  and  100  kilograms  when  the  deflections  were  found  to  be 
6.6  and  13.0  millimeters.  Show  that  the  mean  value  of  E  was 
74  500  kilos  per  square  centimeter. 


ART.  36.    COMPARATIVE  DEFLECTION  AND  STIFFNESS. 

From  the  two  preceding  articles  the  following  values  of  the 
maximum  deflections  may  now  be  written  and  their  compari- 
son will  show  the  relative  stiffness  of  the  different  cases. 

i    wr 

For  a  cantilever  loaded  at  the  end  with  W,        A  =     —  .  — — . 


78  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      ClI.  III. 

i    wr 

For  a  cantilever  uniformly  loaded  with  W,          A  —     -  .  -7^. 

O        ±L1 

i     wr 

For  a  simple  beam  loaded  at  middle  with  W,      A  =  —  .  -—  —  . 

40      rLl 

c       wi* 

For  a  simple  beam  uniformly  loaded  with  W,     4  =  -|--  .  —  —  . 

354     hi 

The  relative  deflections  of  these  four  cases  are  hence  as  the 
numbers  i,  |,  TV,  and  T|^. 

These  equations  also  show  that  the  deflections  vary  directly 
as  the  load,  directly  as  the  cube  of  the  length,  and  inversely  as 

E  and  /.     For  a  rectangular  beam  /  =  --  ,  and  hence  the  de- 

flection of  a  rectangular  beam  is  inversely  as  its  breadth  and 
inversely  as  the  cube  of  its  depth. 

The  stiffness  of  a  beam  is  indicated  by  the  load  that  it  can 
carry  with  a  given  deflection.  From  the  above  it  is  seen  that 
the  value  of  the  load  is, 


, 

where  m  has  the  value  3,  8,  48,  or  -^U-  as  the  case  may  be. 
Therefore,  the  stiffness  of  a  beam  varies  directly  as  E,  directly 
as  /,  and  inversely  as  the  cube  of  its  length,  and  the  relative 
stiffness  of  the  above  four  cases  is  as  the  numbers  i,  2$,  16, 
and  25f  .  From  this  it  appears  that  the  laws  of  stiffness  are 
very  different  from  those  of  strength.  (Art.  29.) 

Prob.  62.  Compare  the  strength  and  stiffness  of  a  joist  3X8 
inches  when  laid  with  flat  side  vertical  and  when  laid  with  nar- 
row side  vertical. 

Prob.  63.  Find  the  thickness  of  a  white  pine  plank  of  8  feet 
span  required  not  to  bend  more  than  ^g-jj-th  of  its  length  under 
a  head  of  water  of  20  feet. 


ART.  37.   RELATION  BETWEEN  DEFLECTION  AND   STRESS.     79 


ART.  37.    RELATION  BETWEEN  DEFLECTION  AND  STRESS. 

Let  the  four  cases  discussed  in  Arts.  29  and  36  be  again  con- 
sidered.    For  the  strength, 


W=  n  — ,     where  n  =  i,  2,  4,  or  8. 


For  the  stiffness, 


EIA 


W  =  m  — — ,     where  m  =  3,  8,  48,  or  76^. 

By  equating  these  values  of  W  the  relation  between  A  and 
is  obtained,  thus, 


mEcA 


or        A  = 


nTS 
mcE  ' 


These  equations,  like  the  general  formula  (4)  and  (5),  are  only 
valid  when  5  is  less  than  the  elastic  limit  of  the  materials. 

This  also  shows  that  the  maximum  deflection  A  varies  as 

P 

-  for  beams  of  the  same  material  under  the  same  unit-stress  5. 

c 

From  the  preceding  articles  the  following  table  may  also  be 
compiled  which  exhibits  the  most  important  results  relating  to 
both  absolute  and  relative  strength  and  stiffness. 


Case. 

Max. 

Vertical 
Shear. 

Max. 
Bending 
Moment. 

Max. 
Stress 
S. 

Max. 
Deflec- 
tion. 

Relative  1 
Strength.  | 

Relative 
Stiffness. 

Cantilever  loaded  at  end, 

W 

Wl 

Wlc 
I 

I  Wl* 
3  ^1 

I 

I 

TT7-7 

i  Wlz 

Cantilever  loaded  uniformly, 

W 

\Wl 

~zT 

2 

2| 

Simple  beam  loaded  at  middle, 

vr. 

tm 

Wlc 

17 

i    Wl* 
48  El 

4 

16 

Simple  beam  loaded  uniformly, 

tw 

\Wl 

Wlc 

17" 

5     WP 
384  ~EI~ 

8 

,51 

SO  CANTILEVER  BEAMS  AND  SIMPLE  BEAMS.      CH.  III. 

Here  the  signs  of  the  maximum  shears  and  moments  are 
omitted  as  only  their  absolute  values  are  needed  in  computa- 
tions. Evidently  the  moments  are  negative  for  the  first  and 
second  cases,  and  positive  for  the  third  and  fourth,  the  direc- 
tion of  the  curvature  being  different. 

Prob.  64.  Find  the  deflection  of  a  wrought  iron  I  heavy  10 
inch  beam  of  9  feet  span  when  strained  by  a  uniform  load  up 
to  the  elastic  limit. 

Prob.  65.  A  wooden  beam  of  breadth  b,  depth  d,  and  span  x 
is  loaded  with  P  at  the  middle.  Find  the  value  of  x  so  that 
rupture  may  occur  under  the  load.  Find  also  the  value  of  x 
so  that  rupture  may  occur  by  shearing  at  the  supports. 

ART.  38.    CANTILEVER  BEAMS  OF  UNIFORM  STRENGTH. 

All  cases  thus  far  discussed  have  been  of  constant  cross- 
section  throughout  their  entire  length.  But  in  the  general 
formula  (4)  the  unit-stress  5  is  proportional  to  the  bending  mo- 
ment M,  and  hence  varies  throughout  the  beam  in  the  same 
way  as  the  moments  vary.  Hence  some  parts  of  the  beam  are 
but  slightly  strained  in  comparison  with  the  dangerous  section. 

A  beam  of  uniform  strength  is  one  so  shaped  that  the  unit- 
stress  5  is  the  same  in  all  fibers  at  the  upper  and  lower  surfaces. 
Hence  to  ascertain  the  form  of  such  a  beam  the  unit-stress  5 

in  (4)  must  be  taken  as  constant  and  —  be  made  to  vary  with 

M.  The  discussion  will  be  given  only  for  the  most  important 
practical  cases,  namely,  those  where  the  sections  are  rectangular. 

For  these  -  equals  —  ,  and  formula  (4)  becomes, 

' 


In  this  bd*  must  vary  with  M  for  forms  of  uniform  strength. 


ART.  38.   CANTILEVER   BEAMS   OF   UNIFORM   STRENGTH.         8l 


For  a  cantilever  beam  with  a  load  P  at  the  end,  M 
and  the  equation  becomes  ^Sbd*  =  Px,  in  which  P  and 
constant.  If  the  breadth  be  taken  as  con- 
stant, d*  varies  with  x  and  the  profile  is 
that  of  a  parabola  whose  vertex  is  at  the 
load,  as  shown  in  Fig.  29.  The  equation  of 

the  parabola  is  d*  =  —  x  from  which  d  may 

kj  O 


Px 
are 


be  found  for  given  values  of  x.     The  walk-  Fig- 29> 

ing  beam  of  an  engine  is  often  made  approximately  of  this 

shape.     If  the  depth  of  the  cantilever  beam 

be  constant  then  b  varies  directly  as  x  and 

hence  the  plan  should  be  a  triangle  as  shown 

in  Fig.  30.     The  value  of  b  for  given  values 

of  P,  S,  d,  and  x  may  be  found  from  the 

6Px 

expression  b  =  -^  •  Fig.  30. 

For  a  cantilever  beam  uniformly  loaded  with  w  per  linear 
unit  M  =  ^wxy,   and  the    equation   becomes  ^Sbd*  = 
in  which   w   and    »S   are    known.     If   the 

* -or > 

breadth  be  taken  as  constant  then  d  varies 
as  x  and  the  elevation  is  a  triangle,  as 
in  Fig.  31,  whose  depth  at  any  point  is 


If  however  the  depth  be 


Fig.  31. 


taken    constant,   then    b  =  -|— -  x*  which  is  the  equation  of  a 

parabola  whose  vertex  is  at  the  free  end  of  the  cantilever  and 
whose  axis  is  perpendicular  to  it.  Or  the  equation  may  be 
satisfied  by  two  parabolas  drawn  upon  opposite  sides  of  the 
center  line  as  shown  in  Fig.  32. 

The  vertical  shear  modifies  in  practice  the  shape  of  these 
forms  near  their  ends.     For  instance,  a  cantilever  beam  loaded 


82 


CANTILEVER  BEAMS   AND   SIMPLE   BEAMS.      CH.  III. 


at  the  end  with  P  requires  a  cross-section  at  the  end  equal  to 

p 

-^- where   Sc  is   the   working  shearing   strength.      This  cross- 

section  must  be  preserved  until  a  value  of 
x  is  reached,  where  the  same  value  of  the 
cross-section  is  found  from  the  moment. 


The  deflection  of  a  cantilever  beam  of 
uniform  strength  is  evidently  greater  than 
that  of  one  of  constant  cross-section,  since 
the  unit-stress  5  is  greater  throughout.  In 
any  case  it  may  be  determined  from  the  general  formula 
(5)  by  substituting  for  M  and  /  their  values  in  terms  of  ^in- 
tegrating twice,  determining  the  constants,  and  then  making 
x  equal  to  /  for  the  maximum  value  of  y. 

For  a  cantilever  beam  loaded  at  the  end  and  of  constant 
breadth,  as  in  Fig.  29,  formula  (5)  becomes, 


J*y  _  i2/!f  _  2       /  S*b 
lx*      EM*      E*y  6Px' 


Integrating  this  twice  and  determining  the  constants,  as  in 
Art.  34,  the  equation  of  the  elastic  curve  is  found  to  be, 


In  this  make  x  =  /,  and  substitute  for  5  its  value  7-75  ,  where 

bal 
d^  is  the  depth  of  the  wall.     Then, 


Ebd?  ' 
which  is  double  that  of  a  cantilever  beam  of  uniform  depth  d. 

For  a  cantilever  beam  loaded  at  the  end  and  of  constant 
depth,  formula  (5)  becomes, 

2S 


ART.  39.      SIMPLE   BEAMS   OF  UNIFORM   STRENGTH.  83 

By  integrating  this  twice  and  determining  the  constants  as 
before,  the  equation  of  the  elastic  curve  is  found,  from  which 
the  deflection  is,  f\pp 


which  is  fifty  per  cent  greater  than  for  one  of  uniform  section. 

Prob.  66.  A  cast  iron  cantilever  beam  of  uniform  strength  is 
to  be  4  feet  long,  3  inches  in  breadth  and  to  carry  a  load  of 
15  ooo  pounds  at  the  end.  Find  the  proper  depths  for  every 
foot  in  length,  using  3  ooo  pounds  per  square  inch  for  the  hori- 
zontal unit-stress,  and  4000  pounds  per  square  inch  for  the 
shearing  unit-stress. 


ART.  39.    SIMPLE  BEAMS  OF  UNIFORM  STRENGTH. 

In  the  same  manner  it  is  easy  to  deduce  the  forms  of  uni- 
form strength  for  simple  beams  of  rectangular  cross-section. 

For  a  load  at  the  middle  and  breadth  constant,  M  =  \Px, 

\P 

and  hence,  ±Sbd*  =  %Px.     Hence  d*  =  ^—  *,  from  which  values 

OP 

of  d  may  be  found  for  assumed  values  of  x.  Here  the  profile 
of  the  beam  will  be  parabolic,  the  vertex  being  at  the  support, 
and  the  maximum  depth  under  the  load. 

For  a  load  at  the  middle  and  depth  constant,  M  =  \Px  as 

^P 

before.     Hence  b  =  -jr^*,  and  the  plan  must  be  triangular  or 

lozenge  shaped,  the  width  uniformly  increasing  from  the  sup- 
port to  the  load. 


For  a  uniform  load  and  constant  breadth,  M  =  %wlx  — 
and  hence,  d*=  ^—  (Ix  —  x*),  and  the  profile  of  the  beam  must 
be  elliptical,  or  preferably  a  half-ellipse. 


84  CANTILEVER  BEAMS  AND   SIMPLE  BEAMS.      CH.  III. 

3Z£/ 

For  a  uniform  load  and  constant  depth,  b  =  -|—  (Ix  —  x*)  and 

hence  the  plan  should  be  formed  of  two  parabolas  having  their 
vertices  at  the  middle  of  the  span. 

The  figures  for  these  four  cases  are  purposely  omitted,  in 
order  that  the  student  may  draw  them  for  himself ;  if  any  dif- 
ficulty be  found  in  doing  this  let  numerical  values  be  assigned 
to  the  constant  quantities  in  each  equation,  and  the  variable 
breadth  or  depth  be  computed  for  different  values  of  x. 

In  the  same  manner  as  in  the  last  article,  it  can  be  shown 
that  the  deflection  of  a  simple  beam  of  uniform  strength  loaded 
at  the  middle  is  double  that  of  one  of  constant  cross-section  if 
the  breadth  is  constant,  and  is  one  and  one-half  times  as  much 
if  the  depth  is  constant. 

Prob.  67.  Draw  the  profile  for  a  cast  iron  beam  of  uniform 
strength,  the  span  being  8  feet,  breadth  3  inches,  load  at  the 
middle  30,000  pounds,  using  the  same  working  unit-stresses  as 
in  Prob.  66. 

Prob.  68.  Find  the  deflection  of  a  steel  spring  of  constant 
depth  and  uniform  strength  which  is  6  inches  wide  at  the  mid- 
dle, 52  inches  long,  and  loaded  at  the  middle  with  600  pounds, 
the  depth  being  such  that  the  maximum  fiber  stress  is  20  ooo 
pounds  per  square  inch. 


ART.  40.         BEAMS   OVERHANGING  ONE  SUPPORT. 


85 


CHAPTER   IV. 
RESTRAINED  BEAMS  AND  CONTINUOUS  BEAMS. 

ART.  40.    BEAMS  OVERHANGING  ONE  SUPPORT. 

A  cantilever  beam  has  its  upper  fibers  in  tension  and  the 
lower  in  compression,  while  a  simple  beam  has  its  upper  fibers 
in  compression  and  the  lower  in  tension.  Evidently  a  beam 
overhanging  one  support, 
as  in  Fig.  31,  has  its  over- 
hanging part  in  the  con- 
dition of  a  cantilever, 
and  the  part  near  the 
other  end  in  the  condi- 
tion of  a  simple  beam. 
Hence  there  must  be  a 
point  /  where  the  stresses 
change  from  tension  to 
compression,  and  where 

the     curvature     changes  Fig.  33. 

from  positive  to  negative.  This  point  /  is  called  the  inflection 
point ;  it  is  the  point  where  the  bending  moment  is  zero.  An 
overhanging  beam  is  said  to  be  subject  to  a  restraint  at  the 
support  beyond  which  the  beam  projects,  or,  in  other  words, 
there  is  a  stress  in  the  horizontal  fibers  over  that  support. 

Since  the  beam  has  but  two  supports,  its  reactions  may  be 
found  by  using  the  principle  of  moments  as  in  Art.  16.  Thus, 
if  the  distance  between  the  supports  be  /,  the  length  of  the 


86  RESTRAINED  AND  CONTINUOUS  BEAMS.  CH.  IV. 

overhanging  part  be  m,  and  the  uniform  load  per  linear  foot  be 
w,  the  two  reactions  are, 

„        wl       wn?  „        wl    .  .   wirf 


whose  sum  is  equal  to  the  total  load  wl  -\-wrn.  Here,  as  in 
all  cases  of  uniform  load,  the  lever  arms  are  taken  to  the  cen- 
ters of  gravity  of  the  portions  considered. 

When  the  reactions  have  been  found,  the  vertical  shear  at 
any  section  can  be  computed  by  Art.  17,  and  the  bending  mo- 
ment by  Art.  18,  bearing  in  mind  that  for  a  section  beyond  the 
right  support  the  reaction  R^  must  be  considered  as  a  force 
acting  upward.  Thus,  for  any  section  distant  x  from  the  left 
support, 

When  x  is  less  than  /,  When  x  is  greater  than  /, 

t-  wx, 


The  curves  corresponding  to  these  equations  are  shown  on 
Fig.  33.    The  shear  curve  consists  bf  two  straight  lines;   V=  R, 

r> 

when  x  =  o,  and  V  =  o  when  x  •=.  —  -  ;    at  the  right  support 

V  =Rl  —  wl  from  the  first  equation,  and   V  =  Rl  -\-  Ry  —  wl 
from  the  second  ;    V  —  o  when  x  =  l-\-  m.    The  moment  curve 

consists  of  two  parts  of  parabolas  ;  M  =  o  when  x  =  o,  M  is  a 
•p 

maximum  when  x  =  —  -,  M  =  o  at  the  inflection  point  where 

w 

x  =  -  i,  M  has  its  negative  maximum  when  x  =  /,  and  M=o 

w 

when  x  —  I  -\-  m.     The  diagrams  show  clearly  the  distribution 
of  shears  and  moments  throughout  the  beam. 

For  example,  if  /=  20  feet,  m  =  10  feet,  and  w  =  40  pounds 
per  linear  foot,  the  reactions  are  R^  —  30x3  and  R^  =  900  pounds. 


ART.  40.         BEAMS  OVERHANGING  ONE  SUPPORT.  87 

Then  the  point  of  zero  shear  or  maximum  moment  is  at 
x  =  7.5  feet,  the  inflection  point  at  x  —  15  feet,  the  maximum 
shears  are  -j-  300,  —  500,  and  -f-  400  pounds,  and  the  maximum 
bending  moments  are  -f-  II25  and  —  2  ooo  pound-feet.  Here 
the  negative  bending  moment  at  the  right  support  is  numeri- 
cally greater  than  the  maximum  positive  moment.  The  rela- 
tive values  of  the  two  maximum  moments  depend  on  the  ratio 
of  m  to  /;  if  m  =  o  there  is  no  overhanging  part  and  the  beam 
is  a  simple  one  ;  if  m  =  \l  the  case  is  that  just  discussed  ;  if 
m  =  I  the  reaction  R^  is  zero,  and  each  part  is  a  cantilever  beam. 

After  having  thus  found  the  maximum  values  of  Fand  M 
the  beam  may  be  investigated  by  the  application  of  formulas 
(3)  and  (4)  in  the  same  manner  as  a  cantilever  or  simple  beam. 
By'the  use  of  formula  (5)  the  equation  of  the  elastic  curve  be- 
tween the  two  supports  is  found  to  be, 


From  this  the  maximum  deflection  for  any  particular  case  may 

dy 
be  determined  by  putting  -~  equal  to  zero,  solving  for  x,  and 

then  finding  the  corresponding  value  of  y. 

If  concentrated  loads  be  placed  at  given  positions  on  the 
beam  the  reactions  are  found  by  the  principle  of  moments,  and 
then  the  entire  investigation  can  be  made  by  the  methods  above 
described. 

Prob.  69.  Three  men  carry  a  stick  of  timber,  one  taking  hold 
at  one  end  and  the  other  two  at  a  common  point.  Where 
should  this  point  be  so  that  each  may  bear  one  third  the 
weight  ?  Draw  the  diagrams  of  shears  and  moments. 

Prob.  70.  A  light  1  2-inch  I  beam  25  feet  long  is  used  as  a 
floor  beam  in  a  bridge  with  one  sidewalk,  the  distance  between 
the  supports  being  20  feet.  Find  its  factor  of  safety  when  the 
whole  beam  is  loaded  with  I  200  pounds  per  linear  foot,  and 
also  when  only  the  20  feet  roadway  is  loaded. 


88  RESTRAINED  AND  CONTINUOUS  BEAMS.        CH.  IV. 

ART.  41.    BEAMS  FIXED  AT  ONE  END  AND  SUPPORTED  AT 
THE  OTHER. 

A  beam  is  said  to  be  fixed  at  the  end  when  it  is  so  restrained 

in  a  wall  that  the  tangent  to  the  elastic  curve  at  the  wall  is 

horizontal.     Thus,  in  Fig.  33,  if  the  part  m  is  of  such  a  length 

that  the  tangent  over  the  right  sup- 

—  *--  »  i         i  \     port  is  horizontal,  the  part  /  is  in  the 

same  condition  as  a  beam  fixed  at 
one  end  and  supported  at  the  other. 
Fig.  34  shows  the  practical  arrange- 
ment of  such  a  beam,  the  left  sup- 
Flgg  34-  port  being  upon  the  same  level  as 

the  lower  side  of  the  beam  at  the  wall.  The  reactions  of  such 
a  beam  cannot  be  determined  by  the  principles  of  statics  alone, 
but  the  assistance  of  the  equation  of  the  elastic  curve  must  be 
invoked. 

Case  I.  For  a  uniform  load  over  the  whole  beam,  as  in  Fig. 
34,  let  R  be  the  reaction  at  the  left  end.  Then  for  any  section 
the  bending  moment  is  Rx  —  $wx*.  Hence  the  differential 
equation  of  the  elastic  curve  is, 


Integrate  this  once  and  determine  the  constant  from  the  neces- 
sary condition  that  —  =  o  when  x  =  I.  Integrate  again  and 
find  the  constant  from  the  fact  that  y  =  o  when  x  —  o.  Then, 


Here  also  7=0  when  x  =  /,  and  therefore  R  =  %wl. 

The  moment  at  any  point  now  is  M  =  \wlx  —  favjf,  and  by 
placing  this  equal  to  zero  it  is  seen  that  the  point  of  inflection 
isat^r  =  |/.  By  the  method  of  Art.  24  it  is  found  that  the 


ART.  41.  BEAMS  FIXED  AT   ONE   END.  89 


maximum  moments  are  -{-^-fy-gwl*  and  —  -Jw/2,  and  that  the 
distribution  of  moments  is  as  represented  in  Fig.  34. 

The  point  of  maximum  deflection  is   found  by  placing  — 

equal  to  zero  and  solving  for  x.  Thus  %x*  —  glx*  -f-  /3  =  o, 
one  root  of  which  is  x  =  -\-o.^2i^l,  and  this  inserted  in  the 
value  of  y  gives, 

wts 
A  —  0.0054  ^r, 

for  the  value  of  the  maximum  deflection. 

Case  II.  For  a  load  at  the  middle  it  is  first  necessary  to  con- 
sider that  there  are  two  elastic  curves  having  a  common  ordi- 
nate  and  a  common  tangent  under  the  load,  since  the  expres- 
sions for  the  moment  are  different  on  opposite  sides  of  the 
load.  Thus,  taking  the  origin  as  usual  at  the  supported  end, 

On  the  left  of  the  load, 

(a)  &=£*, 


(b)  EI      =  IRS 

(c)  £fy=$Xx3 

On  the  right  of  the  load  the  similar  equations  are, 


= 

(c)f  Ely  =  %Rx* 

To  determine  the  constants  consider  in  (c)  that  y  —  o  when 
x  =  o  and  hence  that  C3  =  o.  In  (b)1  the  tangent  ^L  =  o  when 
x  —  I  and  hence  Ct  =  —  ±Rl.  Since  the  curves  have  a  common 


90  RESTRAINED  AND   CONTINUOUS  BEAMS.  CH.  IV. 

tangent  under  the  load  (b)  =  (b)'  for  x  =%l,  and  thus  the  value 
of  Cl  is  found.  Since  the  curves  have  a  common  ordinate 
under  the  load  (c)  =  (c)'  when  x  =  £/,  and  thus  Ct  is  found. 

Then, 

Rx*       Prx      Rl*x 
(c}  Ety  =  —  +—-—-, 

RX*    PX*    pix*    Rrx     pr 


From  the  second  of  these  the  value  of  the  reaction  is  R  = 


The  moment  on  the  left  of  the  load  is  now  M  =  -£-$Px,  and 
that  on  the  right  M  =  —  \%Px  -\-  %PL  The  maximum  posi- 
tive moment  obtains  at  the  load  and  its  value  is  -£%Pl.  The 
maximum  negative  moment  occurs  at  the  wall,  and  its  value  is 
-faPl.  The  inflection  point  is  at  x  =  -fal.  The  deflection  un- 
der the  load  is  readily  found  from  (c)  by  making  x  =  £/.  The 
maximum  deflection  occurs  at  a  less  value  of  x,  which  may  be 
found  by  equating  the  first  derivative  to  zero. 

Case  III.  For  a  load  at  any  point  whose  distance  from  the 
left  support  is  kl,  the  following  results  may  be  deduced  by  a 
method  exactly  similar  to  that  of  the  last  case. 

Reaction  at  supported  end        =     %P(2  — 
Reaction  at  fixed  end  =     \P(lk  — 

Maximum  positive  moment      =     \Plk(2.  — 
Maximum  negative  moment     =     %Pl(k  —  k*). 
The  absolute  maximum  deflection  occurs  under  the  load  when 


Prob.  71.  Draw  the  diagrams  of  -shears  and  moments  for 
a  load  at  the  middle,  taking  P=  600  pounds  and  /=  12  feet. 

Prob.  72.  Find  the  position  of  load  Pwhich  gives  the  maxi- 
mum positive  moment.  Find  also  the  position  which  gives  the 
maximum  negative  moment. 


ART.  42.      BEAMS   OVERHANGING  BOTH   SUPPORTS.  9! 

ART.  42.    BEAMS  OVERHANGING  BOTH  SUPPORTS. 
When  a  beam  overhangs  both  supports  the  bending  moments 
for  sections  beyond  the  supports  are  negative,  and  in  general 
between  the  supports  there  will  be  two  inflection  points.     If 
the    lengths    m   and  n    be 
equal  the  reactions  will  be 
equal  under    uniform    load, 
each  being  one  half  of  the 
total    load.      In    any  case, 
whatever  be  the   nature  of 
the  load,J:he  reactions    may  Flg>  35' 

be  found  by  the  principle  of  moments  (Art.  16),  and  then  the 
vertical  shears  and  bending  moments  may  be  deduced  for  all 
sections,  after  which  the  formulas  (3)  and  (4)  can  be  used  for 
any  special  problem. 

Under  a  uniformly  distributed  load,  and  m  =  n,  which  is  the 
most  important  practical  case,  each  reaction  is  wm  -(-  \wl,  the 
maximum  shears  at  the  supports  are  wm  and  fw/,  the  maxi- 
mum moment  at  the  middle  is  -}-  w(£/2  —  fytf),  the  maximum 
moment  at  each  support  is  —  Igivn?,  and  the  inflection  points 
are  distant  \  VT  —  4»«*  from  the  middle  of  the  beam.  Fig.  35 
shows  the  distribution  of  moments  for  this  case.  If  m  =  o,  the 
beam  is  a  simple  one ;  if  /  =  o,  it  consists  of  two  cantilever 
beams. 

Prob.  73.  If  m  =  n  in  Fig.  35,  find  the  ratio  of  /  to  m  in 
order  that  the  maximum  positive  moment  may  numerically 
equal  the  maximum  negative  moment. 

Prob.  74.  A  bridge  with  two  sidewalks  has  a  wooden  floor 
beam  14  X  15  inches  and  30  feet  long,  the  distance  between 
supports  being  20  feet  and  each  sidewalk  5  feet.  Find  its 
factor  of  safety  under  a  uniformly  distributed  load  of  29  ooo 
pounds. 


92  RESTRAINED   AND   CONTINUOUS  BEAMS.  CH.  IV. 

ART.  43.  BEAMS  FIXED  AT  BOTH  ENDS. 
If,  in  Fig.  35,  the  distances  m  and  n  be  such  that  the  elastic 
curve  over  the  supports  is  horizontal  the  central  span  /  is 
said  to  be  a  beam  fixed  at  both  ends.  The  lengths  m  and  n 
which  will  cause  the  curve  to  be  horizontal  at  the  support  can 
be  determined  by  the  help  of  the  elastic  curve.  For  uniform 
load  n  =  m  and  the  bending  moment  at  any  section  in  the 
span  /  distant  x  from  the  left  support  is, 

M  =  (wm  +  \wl~]x  -  %w(m  +  x)\ 
which  reduces  to  the  simpler  form, 


in  which  M'  represents  the  unknown  bending  moment  — 
at  the  left  support. 

Again,  for  a  single  load  P  at  the  middle  of  /in  Fig.  35  the 
elastic  curve  can  be  regarded  as  kept  horizontal  at  the  left  sup- 
port by  a  load  Q  at  the  end  of  the  distance  m.  Then  the  bend- 
ing moment  at  any  section  distant  x  from  the  left  support,  and 
between  that  support  and  the  middle,  is, 


which  reduces  to, 

M=M'  +  lPxt 

in  which  M'  denotes  the  unknown  moment  —  Qm  at  the  left 
support.  The  problem  of  finding  the  bending  moment  at  any 
section  hence  reduces  to  that  of  determining  M'  the  moment 
at  the  support. 

Case  I.  For  a  uniform  load  the  general  equation  of  the  elastic 
curve  now  is, 


ART.  43.  BEAMS   FIXED   AT   BOTH   ENDS.  93 

Integrating  this  twice,  making  -^-  =  o  when  x  =  o  and  also 


when  x  =  /,  the  value  of  M'  , 


is   found   to   be  ---  ,  and 
12 

the   linear  equation   of    the 
elastic  curve  is, 


From  this  the  maximum  deflection  is  found  to  be, 


The  inflection  points  are  located  by  making  M  —  o,  which 
gives  x  =  ^/±  I  A  /  —  .  The  maximum  positive  moment  is  at 

the  middle  and  its  value  is   —  ;    accordingly  the  horizontal 

24 

stress  upon  the  fibers  at  the  middle  of  the  beam  is  one  half 
that  at  the  ends.  The  vertical  shear  at  the  left  end  is  %wl,  at 
the  middle  o,  and  at  the  right  end  —  \wl. 

Case  II.  For  a  load  at  the  middle  the  general  equation  of 
the  elastic  curve  between  the  left  end  and  the  load  is, 


and  in  a  similar  manner  to  that 

of  the  last  case  it  is  easy  to  find 

that  the  maximum  negative  mo-  ^s-  37. 

ments  are  ^Pl,  that  the  maximum  positive  moment  is  -J-/Y,  that 

the  inflection  points  are  half-way  between  the  supports  and  the 

pr 

load,  and  that  the  maximum  deflection  is =^r  • 


Case  III.  For  a  load  P  at  a  distance  kl  at  the  left  end  let 


94  RESTRAINED   AND   CONTINUOUS   BEAMS.  CM.  IV. 

M'  and  V  denote  the  unknown  bending  moment  and  vertical 
shear  at  that  end.  Then  on  the  left  of  the  load, 

M=M'+  V'x, 
and  on  the  right  of  the  load 

M  =  M1  +  V'x  —  P(x  -  kl\ 

By  inserting  these  in  the  general  formula  (5),  integrating  each 
twice  and  establishing  sufficient  conditions  to  determine  the 
unknown  M'  and  V  and  also  the  constants  of  integration,  the 
following  results  may  be  deduced, 

Shear  at  left  end  =  P(\  -  3/P  -f  2/P), 

Shear  at  right  end  =  Pk\^  —  2k). 

Moment  at  left  end  =  —  Plk(\  —2k-}-  &*), 

Moment  at  right  end  =  —  Plk\\  —  k\ 

Moment  under  load  =  -f  Plk\2  —  ^k  -f  2/P). 

If  k  —  \  the  load  is  at  the  middle  and  these  results  reduce  to 
the  values  found  in  Case  II. 

Prob.  75.  Show  from  the  results  above  given  for  Case  III 

that   the   inflection   points  are   at  the   distances  -  -    and 

I       2k 


from  the  left  end. 


3-2* 

Prob.  76.  What  wrought  iron  I  beam  is  required  for  a  span 
of  24  feet  to  support  a  uniform  load  of  40000  pounds,  the  ends 
being  merely  supported?  What  one  is  needed  when  the  ends 
are  fixed  ? 

ART.  44.    COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS. 

As  the  maximum  moments  for  restrained  beams  are  less 
than  for  simple  beams  their  strength  is  relatively  greater.  This 
was  to  be  expected,  since  the  restraint  produces  a  negative 
bending  moment  and  lessens  the  deflection  which  would  other- 


ART.  44.  COMPARISON  OF  RESTRAINED  AND  SIMPLE  BEAMS.  95 


wise  occur.  The  comparative  strength  and  stiffness  of  canti- 
levers and  simple  beams  is  given  in  Art.  37.  To  these  may 
now  be  added  four  cases  from  Arts.  41  and  43,  and  the  follow- 
ing table  be  formed,  in  which  W  represents  the  total  load, 
whether  uniform  or  concentrated. 


Beams  of  Uniform  Cross-section. 

Maximum 
Moment. 

Maximum 
Deflection. 

Relative 
Strength. 

Relative 
Stiffness. 

Cantilever,  load  at  end 

Wl 

i-lr 

I 

I 

Cantilever,  uniform  load 

im 

8  ~EI 

2 

H 

Simple  beam,  load  at  middle 

km 

i    Wl* 

48  ~EI 

4 

16 

e       Wfi 

Simple  beam  uniformly  loaded 

\m 

5      **  i 
384  ~EI 

8 

25f 

Beam  fixed  at  one  end,  supported  at 
other,  load  near  middle 

0.192  Wl 

Wl* 
0.0182— 

5-2 

18.3 

Beam  fixed  at  one  end,  supported  at 

Wl* 

other,  uniform  load 

\Wl 

0.0054-^ 

8 

62 

Beam   fixed   at   both    ends,    load  at 

Wl* 
I       Wl 

middle 

i  wi 

8 

64 

192   El 

Beam   fixed   at    both   ends,   uniform 

load 

&WI 

384^7 

12 

128 

This  table  shows  that  a  beam  fixed  at  both  ends  and  uni- 
formly loaded  is  one  and  one-half  times  as  strong  and  five  times 
as  stiff  as  a  simple  beam  under  the  same  load.  The  advantage 
of  fixing  the  ends  is  hence  very  great. 

Prob.  77.  Prove,  for  a  uniformly  loaded  beam  with  equal 
overhanging  ends,  that  the  deflection  at  the  middle  is  given  by 


the  formula 


'  -  24wa). 


384/i/ 

Prob.  78.  Find  the  deflection  of  a  I  Q-inch  beam  of  6  feet 
span  and  fixed  ends  when  loaded  at  the  middle  so  that  the 
tensile  and  compressive  stresses  at  the  dangerous  section  are 
14000  pounds  per  square  inch. 


96  RESTRAINED  AND  CONTINUOUS  BEAMS.        CH.  IV. 

ART.  45.    GENERAL  PRINCIPLES  OF  CONTINUITY. 

A  continuous  beam  is  one  supported  upon  several  points  in 
the  same  horizontal  plane.  A  simple  beam  may  be  regarded 
as  a  particular  case  of  a  continuous  beam  where  the  number  of 
supports  is  two.  The  ends  of  a  continuous  beam  are  said  to 
be  free  when  they  overhang,  supported  when  they  merely  rest 
on  abutments,  and  restrained  when  they  are  horizontally  fixed 
in  walls. 

The  general  principles  of  the  preceding  chapter  hold  good 
for  all  kinds  of  beams.  If  a  plane  be  imagined  to  cut  any 
beam  at  any  point  the  laws  of  Arts.  19  and  20  apply  to  the 
stresses  in  that  section.  The  resisting  shear  and  the  resisting 
moment  for  that  section  have  the  values  deduced  in  Art.  21 
and  the  two  fundamental  formulas  for  investigation  are, 

(3)  ScA  =  V, 

(4)  *L=M 

Here  .S,  is  the  vertical  shearing  unit-stress  in  the  section,  and 
5  is  the  horizontal  tensile  or  compressive  unit-stress  on  the 
fiber  most  remote  from  the  neutral  axis  ;  c  is  the  shortest  dis- 
tance from  that  fiber  to  that  axis ;  /  the  moment  of  inertia, 
and  A  the  area  of  the  cross-section.  V  is  the  vertical  shear  of 
the  external  forces  on  the  left  of  the  section,  and  M  is  the 
bending  moment  of  those  forces  with  reference  to  a  point  in 
the  section.  For  any  given  beam  evidently  5S  and  .S  may  be 
found  for  any  section  as  soon  as  V  and  M  are  known. 

The  general  equation  of  the  elastic  line,  deduced  in  Art.  33, 
is  also  valid  for  all  kinds  of  beams.  It  is, 

(0  d'/_M_ 

«'  ~dx>      El 


ART.  45.       GENERAL  PRINCIPLES  OF   CONTINUITY.  97 

where  x  is  the  abscissa  and  y  the  ordinate  of  any  point  of  the 
elastic  curve,  M  being  the  bending  moment  for  that  section, 
and  E  the  coefficient  of  elasticity  of  the  material. 

The  vertical  shear  V  is  the  algebraic  sum  of  the  external 
forces  on  the  left  of  the  section,  or,  as  in  Art.  17, 
V—  Reactions  on  left  of  section  minus  loads  on  left  of  section. 
For  simple  beams  and  cantilevers  the  determination  of  V 
for  any  special  case  was  easy,  as  the  left  reaction  could  be 
readily  found  for  any  given  loads.  For  continuous  beams, 
however,  it  is  not,  in  general,  easy  to  find  the  reactions,  and 
hence  a  different  method  of  determining  V  is  necessary.  Let 
Fig.  37  represent  one  span  of  a  continuous  beam.  Let  Fbe 

the  vertical  shear  for  any  ^ ^ ^ 

section  at  the  distance  x  ^ ^--J,PI      j 

from  the  left  support,  and   )        i  j  \ 

V  the  vertical  shear  at  a  jp~ 

section  infinitely  near  to  Fig.  38. 

the  left  support.  Also  let  2Pl  denote  the  sum  of  all  the  con- 
centrated loads  on  the  distance  x,  and  wx  the  uniform  load. 
Then  because  V  is  the  algebraic  sum  of  all  the  vertical  forces 
on  its  left,  the  definition  of  vertical  shear  gives, 

(6)  V=  V  —  wx  —  2Pt. 

Hence  Fcan  be  determined  as  soon  as  V  is  known. 

The  bending  moment  M  is  the  algebraic  sum  of  the  mo- 
ments of  the  external  forces  on  the  left  of  the  section  with  ref- 
erence to  a  point  in  that  section,  or,  as  in  Art.  18, 

M  =  moments  of  reactions  minus  moments  of  loads. 

For  the  reason  just  mentioned  it  is  in  general  necessary  to  de- 
termine M  for  continuous  and  restrained  beams  by  a  different 
method.  Let  M'  denote  the  bending  moment  at  the  left  sup- 
port of  any  span  as  in  Fig.  30,  and  M"  that  at  the  right  sup- 


98  RESTRAINED  AND   CONTINUOUS  BEAMS.  CH.  IV. 

port,  while  M  is  the  bending  moment  for  any  section  distant  x 
from  the  left  support.  Let  Pl  be  any  concentrated  load  upon 
the  space  x  at  a  distance  >£/  from  the  left  support,  k  being  a 
fraction  less  than  unity,  and  let  w  be  the  uniform  load  per  lin- 
ear unit.  Let  V  be  the  resultant  of  all  the  vertical  forces  on 
the  left  of  a  section  in  the  given  span  infinitely  near  to  the 
left  support,  and  let  m  be  the  distance  of  the  point  of  applica- 
tion of  that  resultant  from  that  support,  Then  the  definition 
of  bending  moment  gives, 

M= 


But  V'm  is  the    unknown   bending   moment  M'  at   the    left 
support.     Hence 

(7)        M  =  M'  +  V'x  -  \wx*  -2.P,(x  -kl\ 

from  which  M  may  be  found  for  any  section  as  soon  as  M'  and 
V  have  been  determined. 

The  vertical  shear  V  at  the  support  may  be  easily  found  if 
the  bending  moments  M'  and  M"  be  known.  Thus  in  equa- 
tion (7)  make  x  =  /,  then  M  becomes  M",  and  hence, 


The  whole  problem  of  the  discussion  of  restrained  and  con- 
tinuous beams  hence  consists  in  the  determination  of  the  bend- 
ing moments  at  the  supports.  When  these  are  known  the 
values  of  M  and  Fmay  be  determined  for  every  section,  and 
the  general  formulas  (3),  (4),  and  (5)  be  applied  as  in  Chapter 
III,  to  the  investigation  of  questions  of  strength  and  deflec- 
tion. The  formulas  (6),  (7),  and  (8)  apply  to  cantilever  and 
simple  beams  also.  For  a  simple  beam  M'  =  M"  =  o,  and 
V  =  R-  For  a  cantilever  beam  M1  =  o  for  the  free  end,  and 
M"  is  the  moment  at  the  wall. 

The  relation  between  the  bending  moment  and  the  vertical 


ART.  46.         PROPERTIES  OF  CONTINUOUS  BEAMS.  99 

shear  at  any  section  is  interesting  and  important.  At  the  sec- 
tion x  the  moment  is  M  and  the  shear  is  V.  At  the  next  con- 
secutive section  x  -j-  dx  the  moment  is  M ' -\-  dM,  which  may 
also  be  expressed  by  M-\-  Vdx.  Hence, 

v=dM 

dx 

This  may  be  proved  otherwise  by  differentiating  (7)  and  com- 
paring with  (6).  From  this  it  is  seen  that  the  maximum  mo- 
ments occur  at  the  sections  where  the  shear  passes  through 
zero. 

Prob.  79.  A  bar  of  length  2/  and  weighing  w  per  linear 
unit  is  supported  at  the  middle.  Apply  formulas  (6)  and  (7) 
to  the  statement  of  general  expressions  for  the  moment  and 
shear  at  any  section  on  the  left  of  the  support,  and  also  at  any 
section  on  the  right  of  the  support. 

ART.  46.    PROPERTIES  OF  CONTINUOUS  BEAMS. 

The  theory  of  continuous  beams  presented  in  the  following 
pages  includes  only  those  with  constant  cross-section  having 
the  supports  on  the  same  level,  as  only  such  are  used  in  engin- 
eering constructions.  Unless  otherwise  stated,  the  ends  will 
be  supposed  to  simply  rest  upon  their  supports,  so  that  there 
can  be  no  moments  at  those  points.  Then  the  end  spans  are 


somewhat  in  the 

ft 

Is                       Is                   I* 

...             r  i  

1 

a  beam  with  one     i 

ovprhancrino-pnrl        j/flf 

^ 

fflifflitov     i 

&a             Qi             Qs 
t                 i 

!           ; 

and   the   other 
spans   somewhat 

W 

Fig.  39- 

in  the  condition  of  a  beam  with  two  overhanging  ends.  At 
each  intermediate  support  there  is  a  negative  moment,  and  the 
distribution  of  moments  throughout  the  beam  will  be  as  repre- 
sented in  Fig.  39. 


IOO  RESTRAINED  AND   CONTINUOUS   BEAMS.  CH.  IV. 

As  shown  in  Art.  45,  the  investigation  of  a  continuous  beam 
depends  upon  the  determination  of  the  bending  moments  at 
the  supports.  In  the  case  of  Fig.  39  these  moments  being 
those  at  the  supports  2,  3,  and  4,  may  be  designated  Mt  ,  M3  , 
and  MI  .  Let  F,  ,  Fa  ,  Vt  ,  and  F4  denote  the  vertical  shear  at  the 
right  of  each  support.  The  first  step  is  to  find  the  moments 
M3,M3,  and  M,  .  Then  from  formula  (8)  the  values  of  V,  ,  V^ 
Vt  ,  and  F4  are  found,  and  thus  by  formula  (7)  an  expression  for 
the  bending  moment  in  each  span  may  be  written,  from  which 
the  maximum  positive  moments  may  be  determined.  Lastly, 
by  formulas  (3)  and  (4)  the  strength  of  the  beam  may  be  in- 
vestigated, and  by  (5)  its  deflection  at  any  point  be  deduced. 

For  example,  let'  the  beam  in  Fig.  39  be  regarded  as  of  four 
equal  spans  and  uniformly  loaded  with  w  pounds  per  linear 
unit.  By  a  method  to  be  explained  in  the  following  articles  it 
may  be  shown  that  the  bending  moments  at  the  supports  are, 


From  formula  (8)  the  vertical  shears  at  the  right  of  the  several 
supports  are, 

F;  =  ttw/,         Vt  =  ftwl,         Vt  =  ftwl,      etc. 
And  from  (6)  those  on  the  left  of  the  supports  2,  3,  4,  etc.,  are 
found  to  be,  —  ffiwl,  —  %\ivl,  —  %^wl,  etc.     From  formula  (7) 
the  general  expressions  for  the  bending  moments  now  are, 

For  first  span,         M  = 

For  second  span,    M=  — 

For  third  span,       M=  —  ^wl*  +  %%wlx  — 

For  fourth  span,    M=  —  ^wl*  +  ftwlx  — 
From   each  of  these  equations  the  inflection   points  may  be 
found  by  putting  M=o,  and  the  point  of  maximum  positive 

moment    by  putting    -—  =  o.     The  maximum  positive   mo- 


ART.  46.         PROPERTIES   OF  CONTINUOUS  BEAMS.  IOI 

ments  are  found  to  have  the  following  values, 

TV\W>         Tfte*''.         TU^\         and        tffavl*. 
For  any  particular  case  the  beam  may  now  be  investigated  by 
formulas  (3)  and  (4). 

The  reactions  at  the  supports  are  not  usually  needed  in  the 
discussion  of  continuous  beams,  but  if  required  they  may 
easily  be  found  from  the  adjacent  shears.  Thus  for  the  above 
case, 


=  |f  wl, 

R,  =££«//+  ifw/  =  ffw/,  etc., 
and  the  sum  of  these  is  equal  to  the  total  load  ^wl. 

The  equation  of  the  elastic  curve  in  any  span  is  deduced  by 
inserting  in  (5)  for  M  its  value  and  integrating  twice.  When 

dy 
x  =  o,  the  tangent  ~-  is  the  tangent  of  the  inclination  at  the 

left  support,  and  when  x  =  I  it  is  the  tangent  of  the  inclination 
at  the  right  support.  When  x  =  o,  and  also  when  x  =  /,  the 
ordinate  y  =  o,  and  from  these  conditions  the  two  unknown 
tangents  may  be  found.  In  general  the  maximum  deflection 
in  any  span  of  a  continuous  beam  will  be  found  intermediate 
in  value  between  those  of  a  simple  beam  and  a  restrained 
beam. 

In  the  following  pages  continuous  beams  will  only  be  inves- 
tigated for  the  case  of  uniform  load.  The  lengths  of  the  spans 
however  may  be  equal  or  unequal,  and  the  load  per  linear  foot 
may  vary  in  the  different  spans. 

Prob.  80.  In  a  continuous  beam  of  three  equal  spans  the 
negative  bending  moments  at  the  supports  are  •fawl*.  Find 
the  inflection  points,  the  maximum  positive  moments  and  the 
reactions  of  the  supports. 


102  RESTRAINED  AND   CONTINUOUS  BEAMS.        CH.  IV. 

ART.  47.    THE  THEOREM  OF  THREE  MOMENTS. 

Let  the  figure  represent  any  two  adjacent  spans  of  a  continu- 
ous beam  whose  lengths  are  /'  and  I"  and  whose  uniform  loads 
per  linear  foot  are  w'  and  w"  respectively.  Let  M',  M",  and 

M'"     represent     the 

^^^^^^5%^%%%%%%^ '       *"       Of'"       tnree     unknown    mo- 

i— » — i    7    •    '  — - -^ 1     ments  at  the  supports. 

&*v>  ZSFF7'  s          Let  F  and  V"  be  the 

vertical  shears  at  the 

Fig- 4a  right  of  the  first  and 

second  supports.     Then,  for  any  section  distant  x  from  the 
left  support  in  the  first  span,  the  moment  is, 


If  this  be  inserted  in  the  general  formula  (5)  and  integrated 
twice  and  the  constants  determined  by  the  condition  that 
y  =  o  when  x  =  o  and  also  when  x  =  /,  the  value  of  the  tan- 
gent of  the  angle  which  the  tangent  to  the  elastic  curve  at  any 
section  in  the  first  span  makes  with  the  horizontal  is  found 
to  be, 

dy  _  \2M'(2x  -  /Q  +  4V'(^  -  n  ~  w'(V?  -  I"} 
dx  24EI 

Similarly  if  the  origin  be  taken  at  the  next  support  the  value 
of  the  tangent  of  inclination  at  any  point  in  the  second  span  is, 

dy_  _  \2M"(2x  -  I" 


dx  ~  24EI 

Evidently  the  two  curves  must  have  a  common  tangent  at  the 
support.  Hence  make  x  —  I'  in  the  first  of  these  and  x  =  o 
in  the  second  and  equate  the  results,  giving, 

=  -  \2M"l"  -V"l"wnl"\ 


ART.  48.    CONTINUOUS  BEAMS  WITH  EQUAL  SPANS.         103 

Let  the  values  of  V  and  V"  be  expressed  by  (8)  in  terms  of 
Mf,  M",  and  M'",  and  the  equation  reduces  to, 


(9)    M'l' 

which  is  the  theorem  of  three  moments  for  continuous  beams 
uniformly  loaded. 

If  the  spans  are  all  equal  and  the  load  uniform  throughout, 
this  reduces  to  the  simpler  form, 

M'  +  4M"  +  M'"  =  -—. 

In  any  continuous  beam  of  s  spans  there  are  s  -\-  I  supports 
and  s  —  i  unknown  bending  moments  at  the  supports.  For 
each  of  these  supports  an  equation  of  the  form  of  (9)  may  be 
written  containing  three  unknown  moments.  Thus  there  will 
be  stated  s  —  i  equations  whose  solution  will  furnish  the  values 
of  the  s  —  I  unknown  quantities. 

Prob.  8  1.  A  simple  wooden  beam  one  inch  square  and  15 
inches  long  is  uniformly  loaded  with  100  pounds.  Find  the 
angle  of  inclination  of  the  elastic  curve  at  the  supports. 


ART.  48.    CONTINUOUS  BEAMS  WITH  EQUAL  SPANS. 

Consider  a  continuous  beam  of  five  equal  spans  uniformly 
loaded.  Let  the  supports  beginning  on  the  left  be  numbered 
i,  2,  3,  4,  5,  and  6.  From  the  theorem  of  three  moments  an 
equation  may  be  written  for  each  of  the  supports  2,  3,  4,  and 
5  ;  thus, 


104  RESTRAINED  AND   CONTINUOUS  BEAMS.  CH.  IV. 

Since  the  ends  of  the  beam  rest  on  abutments  without  restraint 
M!  =  J/8  =  o.  Hence  the  four  equations  furnish  the  means 
of  finding  the  four  moments  M9,  M3,  M4,  Mt.  The  solution 
may  be  abridged  by  the  fact  that  Mt  =  Mt,  and  M3  =  M^ 
which  is  evident  from  the  symmetry  of  the  beam.  Hence, 


From  formula  (8)  the  shears  at  the  right  of  the  supports  are, 

etc. 


From  (7)  the  bending  moment  at  any  point  in  any  span  may 
now  be  found  as  in  Art.  46,  and  by  (3),  (4),  and  (5)  the  complete 
investigation  of  any  special  case  may  be  effected. 

In  this  way  the  bending  moments  at  the  supports  for  any 
number  of  equal  spans  can  be  deduced.  The  following  tri- 
angular table  shows  their  values  for  spans  as  high  as  seven  in 
number.  In  each  horizontal  line  the  supports  are  represented 
by  squares  in  which  are  placed  the  coefficients  of  —  wT.  For 
example,  in  a  beam  of  3  spans  there  are  four  supports  and  the 
bending  moments  at  those  supports  are  o,  —  -fowl*,  — 
and  o. 

/^  —  7\ 

-     _____  No._ofS.Eans_l_ 
Equal  Spans. 
Moments. 


Fig.  41. 

The  vertical  shears  at  the  supports  are  also  shown  in  the 
following  table  for  any  number  of  spans  up  to  5.  The  space 
representing  a  support  shows  in  its  left-hand  division  the  shear 
on  the  left  of  that  support  and  in  its  right-hand  division 


ART.  48.    CONTINUOUS  BEAMS  WITH  EQUAL  SPANS. 


105 


the  shear  on  the  right.  The  sum  of  the  two  shears  for  any 
support  is,  of  course,  the  reaction  of  that  support.  For  ex- 
ample, in  a  beam  of  five  equal  spans  the  reaction  at  the  second 
support  is 


Fig.  42. 

It  will  be  seen  on  examination  that  the  numbers  in  any 
oblique  column  of  these  tables  follow  a  certain  law  of  increase 
by  which  it  is  possible  to  extend  them,  if  desired,  to  a  greater 
number  of  spans  than  are  here  given. 

As  an  example,  let  it  be  required  to  select  a  I  beam  to  span 
four  openings  of  8  feet  each,  the  load  per  span  being  14000 
pounds  and  the  greatest  horizontal  stress  in  any  fiber  to  be 
12  ooo  pounds  per  square  inch.  The  required  beam  must 
satisfy  formula  (4),  or, 

/_      M 

C~   12  OOO 

where  M  is  the  maximum  moment.  From  the  table  it  is  seen 
that  the  greatest  negative  moment  is  that  at  the  second  sup- 
port, or  ~wr.  The  maximum  positive  moments  are, 

2o 

T71 

For  first  span, 


F2 
For  second  span,    max  M  —  M,  -4 = 

2W 


IO6  RESTRAINED  AND  CONTINUOUS  BEAMS.  CH.  IV. 

The  greatest  value  of  M  is  hence  at  the  second  support.    Then, 

/  =3  X  14000  X  8  X  12  _  I2 
c  28  X  12  ooo 

and  from  the  table  in  Art.  30  it  is  seen  that  a  light  7-inch  beam 
will  be  required. 

Prob.  82.  Draw  the  diagram  of  shears  and  the  diagram  of 
moments  for  the  case  of  three  equal  spans  uniformly  loaded. 

Prob.  83.  Find  what  I  beam  is  required  to  span  three  open- 
ings of  12  feet  each,  the  load  on  each  span  being  6000  pounds, 
and  the  greatest  value  of  5  to  be  12000  pounds  per  square 
inch. 

ART.  49.    CONTINUOUS  BEAMS  WITH  UNEQUAL  SPANS. 

As  the  first  example,  consider  two  spans  whose  lengths  are 
/,,/,,  and  whose  loads  per  linear  unit  are  wt  and  wa  .  The 
theorem  of  three  moments  in  (9)  then  reduces  to, 


and  hence  the  bending  moment  at  the  middle  support  is, 


From  this  the  reaction  at  the  left  support  may  be  found  by  (8) 
and  the  bending  moment  at  any  point  by  (7). 

Next  consider  three  spans  whose  lengths  are  /,,/,,  and  /3  , 
loaded  uniformly  with  wl  ,  wa  ,  w3  .  The  bending  moments  at 
the  second  and  third  supports  are  M^  and  M3.  Then  from  (9), 


and  the  solution  of  these  gives  the  values  of  J/2  and  Ma .  A 
very  common  case  is  that  for  which  /a  =  /,  /,  =  /3  =  nl,  and 
wl  =  wt  =  wa  =  w.  For  this  case  the  solution  gives, 

^  _  ^  _          J  + »"       «^« 
4  ' 


ART.  50.      REMARKS   ON  THE  THEORY  OF  FLEXURE.  IO/ 


Here  if  n  =  i,  these  two  moments  become  —  -nr^/",  as  also 
shown  in  the  last  article. 

Whatever  be  the  lengths  of  the  spans  or  the  intensity  of  the 
loads,  the  theorem  of  three  moments  furnishes  the  means  of 
finding  the  bending  moments  at  the  supports.  Then  from  (8), 
(7),  and  (6)  the  vertical  shears  and  bending  moments  at  every 
section  may  be  computed.  Finally,  if  the  material  be  not 
strained  beyond  its  elastic  limit,  formula  (5)  may  be  used  to 
determine  the  deflection,  while  (4)  investigates  the  strength  of 
the  beam. 

Prob.  84.  A  continuous  beam  of  three  equal  spans  is  loaded 
only  in  the  middle  span.  Find  the  reactions  of  the  end  sup- 
ports due  to  this  load. 

Prob.  85.  A  heavy  12-inch  I  beam  of  36  feet  length  covers 
four  openings,  the  two  end  ones  being  each  8  feet  and  the 
others  each  10  feet  in  span.  Find  the  maximum  moment  in 
the  beam.  Then  determine  the  load  per  linear  foot  so  that  the 
greatest  horizontal  unit-stress  may  be  12  ooo  pounds  per  square 
inch. 

ART.  50.    REMARKS  ON  THE  THEORY  OF  FLEXURE. 

The  theory  of  flexure  presented  in  this  and  the  preceding 
chapter  is  called  the  common  theory,  and  is  the  one  universally 
adopted  for  the  practical  investigation  of  beams.  It  should  not 
be  forgotten,  however,  that  the  axioms  and  laws  upon  which 
it  is  founded  are  only  approximate  and  not  of  an  exact  nature 
like  those  of  mathematics.  Laws  (A}  and  (B]  for  instance  are 
true  as  approximate  laws  of  experiment,  but  probably  not  as 
exact  laws  of  science.  Law  (G)  indeed  rests  upon  so  slight 
experimental  evidence  that  it  is  more  of  a  hypothesis  than  an 
established  truth.  Objections  may  also  be  raised  against  the 
formula  (3)  which  supposes  the  vertical  shear  to  be  uniformly 
distributed  over  the  cross-section. 


108  RESTRAINED  AND   CONTINUOUS  BEAMS.  Cil.  IV. 

When  experiments  on  beams  are  carried  to  the  point  of  rup- 
ture and  the  longitudinal  unit-stress  5  computed  from  formula 
(4)  a  disagreement  of  that  value  with  those  found  by  direct 
experiments  on  tension  or  compression  is  observed.  This  is 
often  regarded  as  an  objection  to  the  common  theory  of  flexure, 
but  it  is  in  reality  no  objection,  since  law  (G)  and  formula  (4) 
are  only  true  provided  the  elastic  limit  of  the  material  be  not 
exceeded.  Experiments  on  the  deflection  of  beams  furnish  on 
the  other  hand  the  most  satisfactory  confirmation  of  the  theory. 
When  E  is  known  by  tensile  or  compressive  tests  the  formulas 
for  deflection  are  found  to  give  values  closely  agreeing  with 
those  observed.  Indeed  so  reliable  are  these  formulas  that  it 
is  not  uncommon  to  use  them  for  the  purpose  of  computing  E 
from  experiments  on  beams.  If  however  the  elastic  limit  of 
the  material  be  exceeded,  the  computed  and  observed  deflec- 
tions fail  to  agree. 

On  the  whole  it  may  be  concluded  that  the  common  theory 
of  flexure  is  entirely  satisfactory  and  sufficient  for  the  investi- 
gation of  all  practical  questions  relating  to  the  strength  and 
stiffness  of  beams.  The  actual  distribution  of  the  internal 
stresses  is  however  a  matter  of  very  much  interest  and  this  will 
be  discussed  at  some  length  in  Chapter  VIII. 

The  theory  of  flexure  is  here  applied  to  continuous  beams 
only  for  the  case  of  uniform  loads.  It  should  be  said  however 
that  there  is  no  difficulty  in  extending  it  to  the  case  of  concen- 
trated loads.  By  a  course  of  reasoning  similar  to  that  of  Art. 
48  it  may  be  shown  that  the  theorem  of  three  moments  for 
single  loads  is, 

M'l'  +  2M"(l'  +  /")  -f  M'"l"  =  —  P'l'\k  -  /i-3) 

-  P" 'I '"  (2k  -3/P-f/P). 

Here  as  in  Fig.  37  the  moments  at  three  consecutive  supports 
are  designated  by  M',  M",  and  M'"  and  the  lengths  of  the  two 
spans  by  /'  and  I".  P'  is  any  load  on  the  first  span  at  a  dis- 


ART.  50.      REMARKS   ON   THE  THEORY   OF  FLEXURE.  109 

tance  kl'  from  the  left  support  and  P"  any  load  on  the  second 
span  at  a  distance  kl"  from  the  left  support,  k  being  any  frac- 
tion less  than  unity  and  not  necessarily  the  same  in  the  two 
cases.  From  this  theorem  the  negative  bending  moments  at 
the  supports  for  any  concentrated  loads  may  be  found,  and  the 
beam  be  then  investigated  by  formulas  (6)  and  (4).  For  ex- 
ample, if  a  beam  of  three  equal  spans  be  loaded  with  P  at  the 
middle  of  each  span,  the  negative  moments  at  the  supports 
are  each  ^Pl. 

The  Journal  of  the  Franklin  Institute  for  March  and  April, 
1875,  contains  an  article  by  the  author  in  which  the  law  of  in- 
crease of  the  quantities  in  the  tables  of  Art.  48  is  explained 
and  demonstrated.  A  general  abbreviated  method  of  deduc- 
ing the  moments  at  the  supports  for  both  uniform  and  concen- 
trated loads  on  restrained  and  continuous  beams  is  given  in  ti^e 
Philosophical  Magazine  for  September,  1875.  See  also  Van 
Nostrand's  Science  Series,  No.  25. 

Exercise  4.  Consult  BARLOW'S  Strength  of  Materials  (Lon- 
don, 1837),  and  write  an  essay  concerning  his  experiments  to 
determine  the  laws  of  the  strength  and  stiffness  of  beams.  Con- 
sult also  BALL'S  Experimental  Mechanics. 

Exercise  5.  Consult  Engineering  News,  Vol.  XVIII,  pp.  309, 
352,  404,  443;  Vol.  XIX,  pp.  n,  28,  48,  84;  and  Vol.  XXII, 
p.  121.  Write  an  essay  concerning  certain  erroneous  views  re- 
garding the  theory  of  flexure  which  are  there  discussed.  Con- 
sult also  TODHUNTER'S  History  of  the  Elasticity  and  Strength 
of  Materials. 

Exercise  6.  Procure  six  sticks  of  ash  each  f  X  |  inches  and 
of  lengths  about  8,  12,  and  16  inches.  Devise  and  conduct 
experiments  to  test  the  following  laws :  First,  the  strength  of  a 
beam  varies  directly  as  its  breadth  and  directly  as  the  square 
of  its  depth.  Second,  the  stiffness  of  a  beam  is  directly  as  its 
breadth  and  directly  as  the  cube  of  its  depth.  Third,  a  beam 
fixed  at  the  ends  is  twice  as  strong  and  four  times  as  stiff  as  a 


1 10  RESTRAINED   AND   CONTINUOUS   BEAMS.  CH.  IV. 

simple  beam  when  loaded  at  the  middle.  Write  a  report 
describing  and  discussing  the  experiments. 

Exercise  7.  In  order  to  test  the  theory  of  continuous  beams 
discuss  the  following  experiments  by  FRANCIS  and  ascertain 
whether  or  not  the  ratio  of  the  two  observed  deflections  agrees 
with  theory.  "A  frame  was  erected,  giving  4  bearings  in  the 
same  horizontal  plane,  4  feet  apart,  making  3  equal  spans,  each 
bearing  being  furnished  with  a  knife  edge  on  which  the  beam 
was  supported.  Immediately  over  the  bearings  and  secured  to 
the  same  frame  was  fixed  a  straight  edge,  from  which  the  de- 
flections were  measured.  A  bar  of  common  English  refined 
iron,  12  feet  2f  inches  long,  mean  width  1.535  inches,  mean 
depth  0.367  inches,  was  laid  on  the  4  bearings,  and  loaded  at 
the  center  of  each  span  so  as  to  make  the  deflections  the  same, 
the  weight  at  the  middle  span  being  82.84  pounds  and  at  each 
of  the  end  spans  52.00  pounds.  The  deflections  with  these 
weights  were, 

At  the  center  of  the  middle  span  0.281  inches. 

At  center  of  end  spans,  0.275  and  0.284  inches, 

mean  0.280  inches. 

A  piece  3  feet  ii£  inches  long  was  then  cut  from  each  end  of 
the  bar,  leaving  a  bar  4  feet  4|  inches  long,  which  was  replaced 
in  its  former  position  and  loaded  with  the  same  weight  (82.84 
pounds)  as  before,  when  its  deflection  was  found  to  be  1.059 
inches." 

Prob.  86.  A  beam  of  three  spans,  the  center  one  being  /  and 
the  side  ones  «/,  is  loaded  with  P  at  the  middle  of  each  span. 
Find  the  value  of  n  so  that  the  reactions  at  the  end  may  be 
one-fourth  of  the  other  reactions. 

Prob.  87.  Let  a  beam  whose  cross-section  is  an  isosceles  tri- 
angle have  the  base  b  and  the  depth  d.  Prove  that  if  0.13^  be 
cutoff  from  the  vertex  the  remaining  trapezoidal  beam  will 
be  about  9  per  cent  stronger  than  the  triangular  one. 


ART.  51.  CROSS-SECTIONS  OF  COLUMNS.  in 


CHAPTER  V. 
THE  COMPRESSION  OF  COLUMNS. 

ART.  51.    CROSS-SECTIONS  OF  COLUMNS. 

A  column  is  a  prism,  greater  in  length  than  about  ten  times 
its  least  diameter,  which  is  subject  to  compression.  If  the 
prism  be  only  about  four  or  six  times  as  long  as  its  least  diam- 
eter the  case  is  one  of  simple  compression,  the  constants  for 
which  are  given  in  Art.  6.  In  a  case  of  simple  compression 
failure  occurs  by  the  crushing  and  splintering  of  the  material, 
or  by  shearing  in  directions  oblique  to  the  length.  In  the  case 
of  a  column,  however,  failure  is  apt  to  occur  by  a  side  wise 
bending  which  induces  transverse  stresses  and  causes  the  ma- 
terial to  be  highly  strained  under  the  combined  compression 
and  flexure. 

Wooden  columns  are  usually  square  or  round  and  they  may 
be  built  hollow.  Cast  iron  columns  are  usually  round  and  they 
are  often  cast  hollow.  Wrought  iron  columns  are  made  of  a 
great  variety  of  forms.  A  I  may  be  used  as  a  column,  but  they 
are  usually  made  of  three  or  more  different  shape-irons  riveted 
together.  The  Phoenix  column  is  made  by  riveting  together 
flanged  circular  segments  so  as  to  form  a  dosed  cylinder.  It  is 
clear  that  a  square  or  round  section  is  preferable  to  an  unsym- 
metrical  one,  since  then  the  liability  to  bending  is  the  same  in 
all  directions.  For  a  rectangular  section  the  plane  of  flexure 
will  evidently  be  perpendicular  to  the  longer  side  of  the  cross- 
section,  and  in  general  the  plane  of  flexure  will  be  perpendicular 
to  that  axis  of  the  cross-section  for  which  the  moment  of  in- 
ertia is  the  least.  In  designing  a  column  it  is  hence  advisable 


112  THE  COMPRESSION  OF  COLUMNS.  CH.  V. 

that  the  cross-section  should  be  so  arranged  that  the  moments 
of  inertia  about  the  two  principal  rectangular  axes  may  be 
approximately  equal. 

For  instance,  let  it  be  required  to  construct  a  column  with 
two  I  shapes  and  two  plates  as  shown  in  Fig.  43.    The  I  beams 
are  to  be  light   lo-inch  ones  weighing  30 
pounds  per  linear  foot,   and  having   the 
flanges  4.32  inches  wide.     The  plates  are 
to  be  £  inch  thick,  and  it  is  required  to 
find  their  length  x  so  that  the  liability  to 
bending  about  the  two  axes  shown  in  the 
figure  may  .be  the  same.     From  the  table 
Fie-  «•  in  Art.  30  it  is  ascertained  that  the  moment 

of  inertia  /  of  the  beam  about  an  axis  through  its  center  of 
gravity  and  perpendicular  to  the  web  is  150,  while  the  moment 
of  inertia  /'  about  an  axis  through  the  same  point  and  parallel 
to  the  web  is  nearly  8.  Hence,  for  the  axes  shown  in  the 
figure,  the  moments  of  inertia  are, 
For  axis  perpendicular  to  plates, 

2°-~~  +  2  X8  +  2X9x(f-2.i6)'. 
For  axis  parallel  to  plates, 

2*  *2°'5  +  2  X  0.5*  X  5-25'  +  2  X  150. 

Placing  these  two  expressions  equal,  the  value  of  x  is  found  to 
be  between  14  and  14^  inches. 

Prob.  88.  A  column  is  to  be  formed  of  two  light  1 2-inch  eye- 
beams  connected  by  a  lattice  bracing.  Find  the  proper  distance 
between  their  centers,  disregarding  the  moment  of  inertia  of 
the  latticing. 

Prob.  89.  Two  joists  each  2X4  inches  are  to  be  placed  6 
inches  apart  between  their  centers,  and  connected  by  two  others 
each  8  inches  wide  and  x  inches  thick  so  as  to  form  a  closed 
hollow  rectangular  column.  Find  the  proper  value  of  x. 


ART.  52. 


GENERAL   PRINCIPLES. 


ART.  52.    GENERAL  PRINCIPLES. 

If  a  short  prism  of  cross-section  A  be  loaded  with  the  weight 
P,  the  internal  stress  is  to  be  regarded  as  Uniformly  distributed 
over  the  cross-section,  and  hence  the  compressive  unit-stress  Sc 

p 
is  -.-.     But  for  a  long  prism,  or  column,  this  Is  not  the  case; 

A 

P 

while  the  average  unit-stress  is  -r ,  the  stress  in  certain  parts  of 

the  cross-section  may  be  greater  and  upon  others  less  than  this 
value  on  account  of  the  transverse  stresses  due  to  the  sidewise 
flexure.  Hence  in  designing  a  column  the  load  P  must  be 
taken  as  smaller  for  a  long  one  than  for  a  short  one,  since  evi- 
dently the  liability  to  bending  increases  with  the  length. 

Numerous  experiments  on  the  rupture  of  columns  have 
shown  that  the  load  causing  the  rupture  is  approximately  in- 
versely proportional  to  the  square  of  the  length  of  the  column. 
That  is  to  say,  if  there  be  two  columns  of  the  same  material 
and  cross-section  and  one  twice  as  long  as  the  other,  the  long 
one  will  rupture  under  about  one-quarter  the  load  of  the 
short  one. 

The  condition  of  the  ends  of  columns  exerts  a  great  influence 
upon  their  strength.  Class  (a)  includes  those  with  '  round  ends,' 
or  those  in  such  condition  that  they  are  free  to  turn  at  the  ends. 
Class  (c)  includes  those  whose  ends 
are  '  fixed'  or  in  such  condition  that 
the  tangent  to  the  curve  at  the  ends 
always  remains  vertical.  Class  (b) 
includes  those  with  one  end  fixed 
and  the  other  round.  In  architecture 
it  is  rare  that  any  other  than  class  (c) 
is  used.  In  bridge  construction  and 
in  machines,  however,  columns  of  Flg' 44' 

classes  (6)  and  (a)  are  very  common.     It  is  evident  that  class  (c) 


114  THE  COMPRESSION  OF  COLUMNS.  CH.  V. 

is  stronger  than  (6),  and  that  (b)  is  stronger  than  (a),  and  this  is 
confirmed  by  all  experiments.  Fig.  44  is  intended  as  a  sym- 
bolical representation  of  the  three  classes  of  columns,  and  not 
as  showing  how  the  ends  are  rendered  'round'  and  'fixed'  in 
practical  constructions. 

The  theory  of  the  resistance  of  columns  has  not  yet  been 
perfected  like  that  of  beams,  and  accordingly  the  formulas  for 
practical  use  are  largely  of  an  empirical  character.  The  form 
of  the  formulas  however  is  generally  determined  from  certain 
theoretical  considerations,  and  these  will  be  presented  in  the 
following  articles  as  a  basis  for  deducing  the  practical  rules. 

Prob.  90.  A  pillar  formed  of  two  I  beams  each  weighing  93 
pounds  per  yard  is  1  1  inches  square  and  3  feet  long.  What 
load  will  it  carry  with  a  factor  of  safety  of  5  ? 

ART.  53.    EULER'S  FORMULA. 

Consider  a  column  of  cross-section  A  loaded  with  a  weight  P 
IP        under  whose  action  a  certain  small  sidewise  bending 
1          occurs.     Let  the  column  be  round,  or  free  to  turn  at 
both  ends  as  in  Fig.  45.     Take  the  origin  at  the  upper 
end,  and  let  x  be  the  vertical  and  y  the  horizontal  co- 
ordinate of  any  point  of  the  elastic  curve.     The  gen- 
eral equation  (5)  deduced   in  Art.   33,  applies   to  all 
bodies   subject   to   flexure   provided   the  bending   be 
slight  and  the  elastic  limit  of  the  material  be  not  ex- 
ceeded.     For  the   column   the   bending    moment   is 
Fig.  45-     —  Py,  the  negative  sign  being  used  because  the  curve 
is  concave  to  the  axis  of  x  \  hence, 


The  first  integration  of  this  gives, 


ART.  53.  EULER'S  FORMULA.  115 

dy 
But  when  y  =  the  maximum  deflection  A,  the  tangent  —  =  o. 

Hence  C  =  PA"1,  and  by  inversion, 


The  second  integration  now  gives, 

x  =  (-p)  arc  sin  ^  -\-  C'. 

Here  C'  is  o  because  y  =  o  when  x  =  o.     Hence  finally  the 
equation  of  the  elastic  curve  of  the  column  is, 


This  equation  is  that  of  a  sinusoid.     But  also  y  =  o  when 
x  =  L     Hence  if  n  be  an  integer,  l\-pj\  must  equal  nn,  or, 

P=EI% 

which  is  EULER'S  formula  for  the  resistance  of  columns.     This 
reduces  the  equation  of  the  sinusoid  to, 

.    .         x 
y  =  A  sin  rntj. 

The  three  curves  for  n  =  I,  n  =  2,  and  n  =  3  are  shown  in 
Fig.  46.  In  the  first  case  the 
curve  is  entirely  on  one  side  of 
the  axis  of  x,  in  the  second  case 
it  crosses  that  axis  at  the  mid- 
dle, and  in  the  third  case  it 
crosses  at  \l  and  •§/,  the  points 
of  crossing  being  also  inflection 
points  where  the  bending  mo- 
ment is  zero.  Evidently  the 
greatest  deflection  will  occur  for  the  case  where  n  =  i,  and 


Il6  THE  COMPRESSION   OF  COLUMNS.  CH.  V. 

this  is  the  most  dangerous  case.     Hence, 

|    '  (a)  P-™, 

is  EULER'S  formula  for  columns  with  round  ends. 

A  column  with  one  end  fixed  and  the  other  round  is  closely 
represented  by  the  portion  b'b"  of  the  second  case,  V  being 
the  fixed  end  where  the  tangent  to  the  curve  is  vertical.  Here 
n  =  2,  and  the  length  b'b"  is  three-fourths  of  the  entire  length, 
hence, 


is  EULER'S  formula  for  columns  with  one  end  fixed  and  the 
other  round. 

A  column  with  fixed  ends  is  represented  by  the  portion  c'c" 
of  the  case  c.  Here  n  =  3,  and  the  length  c'c"  is  two-thirds  of 
the  entire  length,  hence, 

W  "  =  4 


is  EULER'S  formula  for  columns  with  fixed  ends. 

From  this  investigation  it  appears  that  the  relative  resist- 
ances of  three  columns  of  the  classes  (a),  (b\  and  (c)  are  as  the 
numbers  i,  2\,  and  4,  when  the  lengths  are  the  same,  and  this 
conclusion  is  approximately  verified  by  experiments.  It  also 
appears  that,  if  the  resistance  of  three  columns  of  the  classes 
(a),  (b\  and'  (c)  are  to  be  equal,  their  lengths  must  be  as  the 
numbers  i,  i£,  and  2. 

The  moment  of  inertia  /  in  the  above  formulas  is  taken  about 
a  neutral  axis  of  the  cross-section  perpendicular  to  the  plane  of 
the  flexure,  and  in  general  is  the  least  moment  of  inertia  of 
that  cross-section,  since  the  column  will  bend  in  the  direction 
which  offers  the  least  resistance.  For  a  rectangular  column 


ART.  54.  HODGKINSON'S  FORMULAS.  117 

whose  greatest  side  is  b  and  least  side  d,  the  formulas  may  be 

written, 

„       mn*Ebd* 

P  =  ,         where  m  =  I,  24,  or  4. 

I2/2 

For  a  cylindrical  column  of  diameter  d  the  formulas  are, 

inn*  Ed*  , 

P  = — ,  where  m  =  I,  2\,  or  4. 

Hence  the  strength  of  a  column  varies  directly  as  its  cross-sec- 
tion and  directly  as  the  square  of  its  least  diameter  or  side.  In 
general  if  r  be  the  least  radius  of  gyration  of  the  cross-section 
the  value  of  /is  Ar*  and  the  formula  may  be  written, 

P       mifEf 

—  =  — — ,          where  m—i,  2$,  or  4, 

which  shows  that  P  varies  as  the  square  of  the  ratio  of  r  to  /. 

The  maximum  deflection  A  is  indeterminate,  so  that  the  load 
P,  given  by  EULER'S  formula,  is  merely  the  load  which  causes 
the  column  to  bend.  Practically  the  bending  of  a  column  is 
the  beginning  of  its  failure. 

Prob.  91.  Show  that  EULER'S  formula  for  the  case  of  a  column 

P  *       * 

fixed  at  one  end  and  entirely  free  at  the  other  is  —  = 


A         4/'   ' 


ART.  54.    HODGKINSON'S  FORMULAS. 

EULER'S  formula  gives  valuable  information  regarding  the 
laws  of  flexure  of  columns,  but  is  difficult  of  direct  practical 
application  because  it  indicates  no  relation  between  the  load  P 
and  the  greatest  internal  compressive  unit-stress.  It  shows 
that  the  strength  of  cylindrical  columns  varies  directly  as  the 
fourth  power  of  the  diameter  and  inversely  as  the  square  of 
the  length.  HODGKIXSON  in  his  experiments  observed  that 
this  was  approximately  but  not  exactly  the  case.  He  therefore 


Il8  THE  COMPRESSION  OF  COLUMNS.  CH.  V. 

wrote  for  each  kind  of  columns  the  analogous  expression, 


and  determined  the  constants  a,  /3,  and  6  from  the  results  of 
his  experiments,  thus  producing  empirical  formulas. 

Let  P  be  the  crushing  load  in  gross  tons,  d  the  diameter  of 
the  column  in  inches,  and  /  its  length  in  feet.  Then  HODG- 
KINSON'S  empirical  formulas  are, 

For  solid  cast  iron  cylindrical  columns, 

^3-5 

P  =  14.9-—         for  round  ends, 

^3.5 
P  =  44.2—^         for  flat  ends, 

For  solid  wrought  iron  cylindrical  columns, 

^3-76 
P  =  42-77-          f°r  round  ends; 

i/3.76 

P  =  134-^-        for  flat  ends. 

These  formulas  indicate  that  the  ultimate  strength  of  flat-ended 
columns  is  about  three  times  that  of  round-ended  ones.  The 
experiments  also  showed  that  the  strength  of  a  column  with 
one  end  flat  and  the  other  end  round  is  about  twice  that  of 
one  having  both  ends  round.  HODGKINSON'S  tests  were  made 
upon  small  columns  and  his  formulas  are  not  so  reliable  as 
those  which  will  be  given  in  the  following  articles.  For  small 
cast  iron  columns  however  the  formulas  are  still  valuable. 

By  the  help  of  logarithms  it  is  easy  to  apply  these  formulas 
to  the  discussion  of  given  cases.  Usually  P  will  be  given  and 
d  required,  or  d  be  given  and  P  required.  By  using  assumed 
factors  of  safety  the  proper  size  of  cylindrical  columns  to  carry 
given  loads  may  also  be  determined.  These  formulas,  it  should 
be  remembered,  do  not  apply  to  columns  shorter  than  about 


ART.  55. 


GORDON'S  FORMULA. 


119 


thirty  times  their  least  diameters.     The  word  flat  used  in  this 
Article  is  to  be  regarded  as  equivalent  to  fixed. 

Prob.  92.  A  cast  iron  cylindrical  column  with  flat  ends  is  3 
inches  diameter  and  8  feet  long.  What  load  will  cause  it 
to  fail? 

Prob.  93.  A  cast  iron  cylindrical  column  with  flat  ends  is  to 
be  7  feet  long  and  carry  a  load  of  200  ooo  pounds  with  a  factor 
of  safety  of  6.  Find  the  proper  diameter. 


ART.  55.    GORDON'S  FORMULA. 

The  formulas  of  EULER  are  defective  because  they  contain 
no  constant  indicating  the  working  or  ultimate  compressive 
strength  of  the  material  and  because  they  apply  only  to  long 
columns.  HODGKINSON'S  formulas  are  unsatisfactory  for  simi- 
lar reasons  and  because  they  do  not  well  represent  the  results 
of  later  experiments.  GORDON'S  formula  was  deduced  to 
remedy  these  defects.  It  may  be  established  by 
the  following  considerations. 

Let  the  column  be  of  rectangular  section, 
the  area  being  A,  the  least  side  d,  the  greatest 
side  b,  and  the  length  /.  Let  P  be  the  load 
upon  it.  The  average  compressive  unit-stress 

at   any   section   is    — ,    but   in   consequence   of 

the  sidewise  deflection  this  is  increased  on  the 
concave  side  and  decreased  on  the  convex 
side  by  an  amount  S.  From  the  fundamen- 
tal equation  (4)  the  value  of  5  is  — ,  and  if  A  be 


the  maximum  deflection  the  greatest  value  of 


Fig.   47- 


is  -r-75-.     Now  if  Sc  be  the  total  compressive  unit-stress  on  the 
bd 


120  THE  COMPRESSION   OF  COLUiMNS.  CH.  V, 

p 

concave  side  5  =  Sc  —  —  ,  and  hence, 
A 

P  _  6PA       6PA_ 
~A  ~   bd*  ~  Ad  ' 

Accordingly  the  value  of  5,  is 

P        P   6J 


The  value  of  A  is  unknown,  but  if  the  curve  of  deflection  be  an 
arc  of  a  circle,  which  it  is  very  nearly,  A  equals  approximately 

r 

-—  ,  in  which  R  represents  the  radius  of  curvature  of  the  column. 
OR 

Now,  as  in  Art.  33,  the  value  of  R  for  the  same  unit-stress  5 
varies  directly  as  c  or  as  the  depth  d.  Hence  A  may  be  taken 
as  varying  directly  as  /"  and  inversely  as  d.  Accordingly  if  k 
be  a  number  depending  upon  the  kind  of  material  and  the 
arrangement  of  the  ends  of  the  column,  the  value  of  Sc  may 

be  written, 

P          P  r 


From  this  the  value  of  the  unit-load  —  is, 

A 

P  =        S< 

A  " 


which  is  called  GORDON'S  formula  for  resistance  of  columns. 

The  quantity  k  cannot  be  determined  theoretically.  As  the 
above  reasoning  shows,  its  value  varies  with  the  form  of  cross- 
section  as  well  as  with  the  kind  of  material  and  the  arrange- 
ment of  the  ends  of  the  column.  For  instance,  the  value  of  k 
is  not  the  same  for  a  circular  section  with  diameter  d  as  for  a 
rectangular  section  whose  least  side  is  d.  It  is  however  not 


ART.  55.  GORDON'S  FORMULA.  121 

uncommon  to  find  this  formula  stated  as  applicable  to  any 
cross-section  whose  least  diameter  is  d. 

In  order  to  determine  k  recourse  must  be  had  to  experiments. 
These  are  usually  conducted  by  loading  columns  to  the  point 
of  rupture;  P,  A,  /,  and  d  are  known,  and  thus  the  constants 
Sc  and  k  may  be  computed.  Theoretically  Sc  is  the  ultimate 
compressive  strength  of  the  material  and  the  values  found  for 
it  by  experiments  on  columns  agree  roughly  with  those  deduced 
by  the  direct  crushing  of  short  specimens.  The  value  of  k  is 
always  less  than  unity  and  it  is  subject  to  great  variation,  even 
in  columns  of  the  same  material.  For  a  column  with  round 
ends  k  is  to  be  regarded  four  times  as  great  as  for  a  column 
with  fixed  ends,  since  both  experiment  and  theory  indicate  that 
a  fixed-ended  column  of  length  2/  has  the  same  strength  as  a 
round-ended  column  of  length  /.  Therefore  for  the  ultimate 
strength  of  columns, 

For  fixed  ends,          -j  = — r^-, 


For  round  ends,        —  = 

A 


The  following  values  of  Sc  and  k  were  deduced  by  GORDON 
from  HODGKINSON'S  experiments. 

For  stone  and  brick,  Sc  =  (Art.  6),  k  =  Tr£ir, 

For  timber  (rectangular  sections),          Sc  =  7  200,      k  =  ^¥, 
For  cast  iron  cylinders,  Sc  =  So  ooo,    k  =  -^, 

For  wrought  iron  (rectangular  sections),  Sc  =  36  ooo,  k  =  -5-3^-$, 
These  values  of  Sc  are  in  pounds  per  square  inch,  while  those 
of  k  are  abstract  numbers. 

The  theoretical  deduction  of  the  above  formula  was  first 
made  by  TREDGOLD,  and  it  is  hence  sometimes  referred  to  as 


122  THE  COMPRESSION  OF  COLUMNS.  CH.  V. 

TREDGOLD'S  formula.  The  reasoning  by  which  it  is  deduced 
is  not  entirely  satisfactory,  and  it  often  fails  to  properly  repre- 
sent the  results  of  experiment. 

Prob.  94.  Find  the  values  of  Sc  and  k  from  the  two  follow- 
ing experiments  on  flat-ended  Phoenix  columns.  The  sectional 
area  of  each  column  was  12  square  inches  and  the  exterior  di- 
ameter 8  inches.  The  length  of  the  first  column  was  25  feet, 
and  it  failed  under  a  load  of  420  ooo  pounds.  The  length  of 
the  second  column  was  10  feet,  and  it  failed  under  a  load  of 
478  ooo  pounds. 

ART.  56.    RANKINE'S  FORMULA. 

The  formula  which  seems  to  most  satisfactorily  represent 
the  results  of  experiments  will  now  be  deduced.  It  is  some- 
times called  GORDON'S  formula,  and  occasionally  it  is  referred 
to  as  "  GORDON'S  formula  modified  by  RANKINE,"  but  the  best 
usage  gives  to  it  the  name  of  RANKINE'S  formula.  It  is  simi- 
lar to  the  formula  of  the  last  Article,  but  has  the  advantage  of 
being  applicable  to  any  form  of  cross-section. 

Let  P  be  the  load  on  the  column,  /  its  length,  A  the  area  of 
its  cross-section,  7  the  moment  of  inertia,  and  r  the  radius  of 
gyration  of  that  cross-section  with  reference  to  a  neutral  axis 
perpendicular  to  the  plane  of  flexure,  and  c  the  shortest  dis- 
tance from  that  axis  to  the  remotest  fiber  on  the  concave  side. 
The  average  compressive  unit-stress  on 

p 

any  cross-section  is  — ,  but  in  consequence 
A 

of  the  flexure  this  is  increased  on  the 
concave  side,  and  decreased  on  the  con- 
vex side.  Thus  in  Fig.  48  the  average 

p 
Fi     8  unit-stress  -r  is  represented  by  cd,  but  on 

the  concave  side  this  is  increased  to  aq,  and  on  the  convex 
side  decreased  to  bq.  The  triangles  pdq  and  qdp  represent 


ART.  56.  RANKINE'S  FORMULA.  123 

the  effect  of  the  flexure  exactly  as  in  the  case  of  beams,  pq 
indicating  the  greatest  compressive  and  qp  the  greatest  ten- 
sile  unit-stress  due  to  the  bending.  Let  the  total  maximum 
unit-stress  aq  be  denoted  by  Sc  and  the  part  due  to  the  flexure 
be  denoted  by  5.  Then, 


Now,  from  the  fundamental  formula  (4)  the  flexural  stress  is 
—  ,  where  Mis  the  external  bending  moment,  which  for  a  col- 

umn has  its  greatest  value  when  M  =  PA,  A  being  the  maxi- 
mum deflection.  /=  Ar*  is  the  well-known  relation  between  / 
and  r.  Hence  the  value  of  5  is, 


c-  _          _ 
=  ~7~  ~~A?' 

By  analogy  with  the  theory  of  beams,  as  in  Art.  37,  the  value 

/» 
of  A  may  be  regarded  as  varying  directly  as  —  .     Hence  if  q  be 

a  quantity  depending  upon  the  kind  of  material  and  the  con- 
dition of  the  ends,  the  total  unit-stress  is, 

c       p   ,     P^ 

*-Z+«2?' 

This  may  now  be  written  in  the  usual  form, 
P__       Sc 

do  ^I+/:' 

which  is  RANKINE'S  formula  for  the  investigation  of  columns. 

The  above  reasoning  has  been  without  reference  to  the  ar- 
rangement of  the  ends  of  the  column.  By  Art.  53  it  is  known 
that  a  column  with  round  ends  must  be  one  half  the  length  of 
one  with  fixed  ends  in  order  to  be  of  equal  strength,  and  that 
a  column  with  one  end  fixed  and  the  other  round  must  be 


I24 


THE   COMPRESSION    OF   COLUMNS. 


Cli.  V. 


three  fourths  the  length  of  one  with  fixed  ends  in  order  to  be 
of  equal  strength.  Therefore  if  q  be  the  constant  for  fixed 
ends,  (-f)V  will  be  the  constant  for  one  end  fixed  and  the  other 
round,  and  2*q  will  be  the  constant  for  both  ends  round. 

The  values  of  q  to  be  taken  for  use  in  formula  (10)  for  the 
examples  and  problems  of  this  chapter  may  be  the  following 
rough  values,  unless  otherwise  stated,  while  the  values  of 
the  ultimate  compressive  unit-stress  Sc  will  be  taken  from 
the  table  in  Art.  6. 


Material. 

Both  Ends 
Fixed. 

Fixed  and 
Round. 

Both    Ends 
Round. 

Timber, 

I 
3  ooo 

1.78 
3  ooo 

4 
3  ooo 

Cast  Iron, 

i 
5  ooo 

1.78 
5  ooo 

4 
5  ooo 

Wrought  Iron, 

i 

1.78 

4 

36  ooo 

36  ooo 

36  ooo 

Steel, 

I 

1.78 

4 

25  ooo 

25  ooo 

25  ooo 

The  very  wide  variation  in  the  values  of  q  found  from  differ- 
ent experiments  shows,  however,  that  little  dependence  can  be 
placed  upon  average  results.  In  any  practical  case  of  impor- 
tance an  effort  should  be  made  to  ascertain  values  of  6",.  and  q 
for  the  special  kind  of  columns  on  hand. 

Prob.  95.  Plot  the  curve  represented  by  formula  (10)  for 
cases  of  wrought-iron  columns  with  fixed  and  with  round  ends, 

P  I 

taking  the  values 'of  —  as  ordinates,  and  the  values  of  —  as  ab- 
A  T 

scissas. 

ART.  57.    RADIUS  OF  GYRATION  OF  CROSS-SECTIONS. 
The  radius  of  gyration  of  a  surface  with  reference  to  an  axis 
is  equal  to  the  square  root  of  the  ratio  of  the  moment  of  iner- 


ART.   58.  INVESTIGATION  OF  COLUMNS.  12$ 

tia  of  the  surface  referred  to  the  same  axis  to  the  area  of  the 
figure.  Or  if  r  be  radius  of  gyration,  7  the  moment  of  inertia, 
and  A  the  area  of  the  surface,  then  I =Ar*. 

In  the  investigation  of  columns  by  formula  (10)  the  value  of 
r*  is  required,  r  being  the  least  radius  of  gyration.  These, 
values  are  readily  derived  from  the  expressions  for  the  mo- 
ment of  inertia  given  in  Art.  23,  the  most  common  cases  being 
the  following, 

For  a  rectangle  whose  least  side  is  d, 

For  a  circle  of  diameter  d, 

For  a  triangle  whose  least  altitude  is  d, 

For  a  hollow  square  section, 

For  a  hollow  circular  section, 

For  I  beams  and  other  shapes,  r2  is  found  by  dividing  the  least 
moment  of  inertia  of  the  cross-section  by  the  area  of  that  cross- 
section.  For  instance,  by  the  help  of  the  table  in  Art.  30,  the 
least  value  of  r2  for  a  light  1 2-inch  I  beam  is  found  to  be 
{£?  =  0.87  inches3. 

Prob.  96.  Compute  the  least  radius  of  gyration  for  a  T  iron 
whose  width  is  4  inches,  depth  4  inches,  thickness  of  flange  £ 
inches,  and  thickness  of  stem  ^f  inches. 

ART.  58.  INVESTIGATION  OF  COLUMNS. 
The  investigation  of  a  column  consists  in  determining  the 
maximum  compressive  unit-stress  Sc  from  formula  (10).  The 
values  of  P,  A,  I,  and  r  will  be  known  from  the  data  of  the  given 
case,  and  q  is  known  from  the  results  of  previous  experiments. 
Then,  p 


126  THE   COMPRESSION  OF  COLUMNS.  CH.  V. 

and,  by  comparing  the  computed  value  of  5,  with  the  ultimate 
strength  and  elastic  limit  of  the  material,  the  factor  of  safety 
and  the  degree  of  stability  of  the  column  may  be  inferred. 

For  example,  consider  a  hollow  wooden  column  of  rectangu- 
lar section,  the  outside  dimensions  being  4X5  inches  and  the 
inside  dimensions  3X4  inches.  Let  the  length  be  18  feet, 
the  ends  fixed,  and  the  load  be  540x3  pounds.  Here  P  = 
5400,  A  =  8  square  inches, /=2i6  inches.  From  the  table 
q  —  _i_.  From  Art.  57, 

5X43-4X33_22I 

12  X  8 
Then  the  substitution  of  these  values  gives, 

_       5400/          216  X  216  \ 

Sc  =  — g-  l^i  +  30QOX22J  =  5430  pounds  per  square  inch. 

Here  the  average  unit-stress  is  675  pounds  per  square  inch, 
but  the  flexure  has  increased  that  stress  on  the  concave  side  to 
5  430  pounds  per  square  inch,  so  that  the  factor  of  safety  is 
only  about  \\. 

Prob.  97.  A  cylindrical  wrought  iron  column  with  fixed  ends 
is  12  feet  long,  6.36  inches  in  exterior  diameter,  6.02  inches  in 
interior  diameter,  and  carries  a  load  of  98  ooo  pounds.  Find 
its  factor  of  safety. 

Prob.  98.  A  pine  stick  3X3  inches  and  12  feet  long  is  used 
as  a  column  with  fixed  ends.  Find  its  factor  of  safety  under  a 
load  of  3  ooo  pounds.  If  the  length  be  only  one  foot,  what  is 
the  factor  of  safety? 

ART.  59.    SAFE  LOADS  FOR  COLUMNS. 

To  determine  the  safe  load  for  a  given  column  it  is  necessary 
to  first  assume  the  allowable  working  unit-stress  Sc .  Then  from 
formula  (10)  the  safe  load  is, 


ART.  60.  DESIGNING   OF   COLUMNS.  I2/ 

Here  A,  /,  and  r  are  known  from  the  data  of  the  given  problem 
and  q  is  taken  from  the  table  in  Art.  56. 

For  example,  let  it  be  required  to  determine  the  safe  load 
for  a  fixed-ended  timber  column,  3X3  inches  square  and  12 
feet  long,  so  that  the  greatest  compressive  unit-stress  may  be 
800  pounds  per  square  inch.  From  the  formula, 

800  X  0 
p—  __ _ _  aoout  700  pounds. 

1  +  3  ooo  X  3~a 

A  short  prism  3X3  inches  should  safely  carry  ten  times  this 
load. 

Prob.  99.  Find  the  safe  load  for  a  heavy  wrought  iron  I  of 
15  inches  depth  and  10  feet  length  when  used  as  a  column  with 
fixed  ends,  the  factor  of  safety  being  4. 

Prob.  100.  Find  the  safe  steady  load  for  a  hollow  cast  iron 
column  with  fixed  ends,  the  length  being  18  feet,  outside 
dimensions  4X5  inches,  inside  dimensions  3X4  inches. 


ART.  60.    DESIGNING  OF  COLUMNS. 

When  a  column  is  to  be  selected  or  designed  the  load  to  be 
borne  will  be  known,  as  also  its  length  and  the  condition  of  the 
ends.  A  proper  allowable  unit-stress  Sc  is  assumed,  suitable  for 
the  given  material  under  the  conditions  in  which  it  is  used. 
Then  from  formula  (i)  the  cross-section  of  a  short  column  or 

p 
prism  is  -=-,  and  it  is  certain  that  a  greater  value  of  the  cross- 

O^. 

section  than  this  will  be  required.  Next  assume  a  form  and 
area  A,  find  r\  and  from  the  formula  (10)  compute  Sc.  If  the 
computed  value  agrees  with  the  assumed  value  the  correct  size 
has  been  selected.  If  not,  assume  a  new  area  and  compute  Se 
again,  and  continue  the  process  until  a  proper  agreement  is 
attained. 


128  THE   COMPRESSION   OF   COLUMNS.  CH.  V. 

For  example,  a  hollow  cast  iron  rectangular  column  of  18  feet 
length  is  to  carry  a  load  of  60  ooo  pounds.  Let  the  working 
strength  S£  be  15  ooo  pounds  per  square  inch.  Then  for  a  short 
length  the  area  required  would  be  four  square  inches.  Assume 
then  that  about  6  square  inches  will  be  needed.  Let  the  sec- 
tion be  "square,  the  exterior  dimensions  6x6  inches,  and  the  in- 
terior dimensions  5!  X  5i  inches.  Then  A  =  5.75,  /=  18  X  12, 
P=  60000,  q=  ^5,  r2  =  5.52,  and  from  (10), 

60000 /  :82  X  I22    \ 

Sf  = i  H )  =  about  30  ooo, 

• '.      5-75    V    ^sooox  5-52' 

which  shows  that  the  dimensions  are  much  too  small.  Again 
assume  the  exterior  side  as  6  inches  and  the  interior  as  5  inches. 
Then  A  =  11,  r*  =  5.08,  and 

6oooo/  i82  X  i22    \ 

S<  =  — A1  +  5000X5.08J  =  about  I5  7oa 

As  this  is  very  near  the  required  working  stress,  it  appears  that 
these  dimensions  very  nearly  satisfy  the  imposed  conditions. 

In  many  instances  it  is  possible  to  assume  all  the  dimensions 
of  the  column  except  one,  and  then  after  expressing  A  and  r  in 
terms  of  this  unknown  quantity,  to  introduce  them  into  (10)  and 
solve  the  problem  by  finding  the  root  of  the  equation  thus 
formed.  For  example,  let  it  be  required  to  find  the  size  of  a 
square  wooden  column  with  fixed  ends  and  24  feet  long  to  sus- 
tain a  load  of  100000  pounds  with  a  factor  of  safety  of  10. 

Here  let  x  be  the  unknown  side ;  then  A  —,  x*  and  r"  =  — . 

From  (10), 

,     242  X  i 


x'      v     '   3  ooo  x 
By  reduction  this  becomes, 

8*4  -  i  ooo*2  =  331  776, 
the  solution  of  which  gives  16.6  inches  for  the  side  of  the  column. 


ART.  6l.  EXPERIMENTS   ON   COLUMNS.  129 

Prob.  101.  Find  the  size  of  a  square  wooden  column  with  fixed 
ends  and  12  feet  in  length  to  sustain  a  load  of  100000  pounds 
with  a. factor  of  safety  of  10.  Find  also  its  size  for  round  ends. 

Prob.  102.  A  hollow  cylindrical  cast  iron  column  is  to  be  de- 
signed to  carry  a  load  of  90  ooo  pounds.  Its  length  is  to  be 
12  feet,  its  ends  flat  or  fixed,  its  exterior  diameter  6  inches,  and 
the  allowable  unit-stress  15  ooo  pounds  per  square  inch.  Find 
the  interior  diameter. 

ART.  61.    EXPERIMENTS  ON  COLUMNS. 

It  is  impossible  to  present  here  even  a  summary  of  the  many 
experiments  that  have  been  made  to  determine  the  laws  of  re- 
sistance of  columns.  The  interesting  tests  made  by  CHRISTIE 
in  1883  for  the  Pencoyd  Iron  Works  will  however  be  briefly 
described  on  account  of  their  great  value  and  completeness  as 
regards  wrought  iron  struts,  embracing  angle,  tee,  beam,  and 
channel  sections.  See  Transactions  of  the  American  Society 
of  Civil  Engineers,  April,  1884. 

The  ends  of  the  struts  were  arranged  in  different  methods ; 
first  flat  ends  between  parallel  plates  to  which  the  specimen  was 
in  no  way  connected  ;  second,  fixed  ends,  or  ends  rigidly 
clamped ;  third,  hinged  ends,  or  ends  fitted  to  hemispherical 
balls  and  sockets  or  cylindrical  pins  ;  fourth,  round  ends,  or  ends 
fitted  to  balls  resting  on  flat  plates. 

The  number  of  experiments  was  about  300,  of  which  about 
one-third  were  upon  angles,  and  one-third  upon  tees.  The 
quality  of  the  wrought  iron  was  about  as  follows :  elastic  limit 
32  ooo  pounds  per  square  inch.  Ultimate  tensile  strength 
49600  pounds  per  square  inch,  ultimate  elongation  18  per  cent 
in  8  inches.  The  length  of  the  specimens  varied  from  6  inches 
up  to  16  feet,  and  the  ratio  of  length  to  least  radius  of  gyra- 
tion varied  from  20  to  480.  Each  specimen  was  placed  in  a 
Fairbanks'  testing  machine  of  50  ooo  pounds  capacity  and  the 
power  applied  by  hand  through  a  system  of  gearing  to  two 


130 


THE  COMPRESSION   OF  COLUMNS. 


CH.  V. 


rigidly  parallel  plates  between  which  the  specimen  was  placed 
in  a  vertical  position.  The  pressure  or  load  was  measured  on 
an  ordinary  scale  beam,  pivoted  on  knife  edges  and  carrying 
a  moving  weight  which  registered  the  pressure  automatically. 
At  each  increment  of  5  OOO  pounds,  the  lateral  deflection  of 
the  column  was  measured.  The  load  was  increased  until  failure 
occurred. 
The  following  are  the  combined  average  results  of  these  care- 


Length  divided  by 
Least  Radius  of 
Gyration. 

Flat  Ends. 

Fixed  Ends. 

Hinged  Ends. 

Round  Ends. 

20 

46000 

46000 

46000 

44000 

40 

40000 

40000 

4OOOO 

36500 

60 

36000 

36000 

36000 

30500 

80 

32000 

32  ooo 

31500 

25  ooo 

100 

29  800 

30000 

28000 

20500 

120 

26300 

28000 

24300 

16  500 

140 

23500 

25500 

21  OOO 

12800 

160 

2O  OOO 

23000 

16  500 

9500 

1  80 

16800 

2O  OOO 

12  8OO 

7500 

200 

14500 

17500 

10800 

6000 

220 

12700 

15000 

8  800 

5000 

240 

II  2OO 

13  ooo 

7500 

4300 

260 

9800 

IIOOO 

6  500 

3  800 

280 

8  500 

IOOOO 

5700 

3  200 

300 

7  200 

9000 

5  ooo 

2800 

320 

6000 

8000 

45oo 

2  500 

340 

5  loo 

7000 

4000 

2  IOO 

360 

4300 

6500 

3500 

I  900 

380 

35oo 

5  800 

3000 

1700 

400 

3000 

5  200 

2  5OO 

I  500 

420 

2  500 

4800 

2300 

I  300 

440 

2  2OO 

4300 

2  IOO 

460 

2000 

3800 

1900 

480 

I  gOO 

1800 

ART.  62.  ON   THE  THEORY  OF  COLUMNS.  131 

fully  conducted  experiments.     The  first  column  gives  the  values 

/  P 

of  -,and  the  other  columns  the  value  of  —  or  the  ultimate  load 

7*  A- 

per  square  inch  of  cross-section.  From  these  results  it  will  be 
seen  that  when  the  strut  is  short  there  is  no  practical  difference 
in  the  strength  of  the  four  classes,  and  that  when  the  strut  is 
long  there  is  but  little  difference  between  those  with  flat  and 
hinged  ends.  The  strength  of  the  long  columns  with  fixed 
ends  appears  to  be  about  3^  times  that  of  the  round-ended 


Prob.  103.  Plot  the  above  experiments,  taking  the  values  of 

/  P 

-  as  abscissas  and  those  of  -r  as  ordinates. 
r  A 


ART.  62.    ON  THE  THEORY  OF  COLUMNS. 

It  has  been  already  remarked  that  the  theory  of  columns  is 
in  a  very  incomplete  condition  compared  with  that  of  beams. 
A  satisfactory  formula  for  the  resistance  of  columns  should  be 
of  such  a  nature  that  for  a  short  block  which  fails  by  pure 
crushing  it  would  reduce  to  the  equation  P=  ASC,  while  for  a 
long  strut  which  fails  by  bending  it  would  reduce  to  an  expres- 
sion like  EULER'S.  The  formula  of  RANKINE  conforms  partly 
to  this  requirement,  but  the  fact  that  it  is  impossible  to  deter- 
mine values  of  q  of  general  applicability  indicates  that  q  is  not 
a  constant,  and  that  the  reasoning  by  which  it  is  deduced  is 
faulty.  Nevertheless  RANKINE'S  formula  applies  so  well  to 
columns  of  medium  length  that  it  is  extensively  employed 
in  this  country  in  the  manner  illustrated  in  the  preceding 
articles. 

For  long  columns  EULER'S  formula  often  represents  fairly 
the  results  of  experiments,  and  since  it  contains  /  it  may  be 
adapted  to  any  form  of  cross-section.  Thus  I=Ar*,  and, 


132  THE  COMPRESSION   OF  COLUMNS.  CH.  V. 


For  round  ends, 
For  fixed  ends, 

P 

n^Er* 

A 
P 

r  ' 

A 

/* 

For  wrought  iron  £  equals  about  25  oooooo  pounds  per  square 
inch  and  hence  for  round  ends, 

if    -  =     200,  300,          400, 

p 

—  =  6250,        2800,        1600, 

A 

and  these  agree  well  with  the  experimental  values  given  in  the 
last  Article. 

The  lateral  deflection  of  a  column  is  indeterminate.  For  a 
long  column  which  fails  wholly  by  bending  any  load  less  than 
that  given  by  EULER'S  formula  produces  theoretically  no  lateral 
deflection,  and  if  when  so  loaded  a  slight  lateral  deflection  is 
caused  by  an  applied  horizontal  force,  the  column  springs  back 
to  the  vertical  position  on  the  removal  of  that  force.  If  the 
column  be  loaded  with  P,  equal  in  amount  to  the  value  given 
by  EULER'S  formula,  it  will  not  spring  back  and  recover  the 
vertical  position,  but  will  remain  in  indifferent  equilibrium 
whatever  be  the  amount  of  the  small  deflection.  If  the  load 
be  made  slightly  larger  than  P  the  column  is  in  unstable  equi- 
librium, the  lateral  deflection  increases  and  failure  by  bending 
occurs.  It  is,  however,  probable  that,  for  a  practical  column 
which,  unlike  the  ideal  column,  is  never  perfectly  straight,  an 
applied  load  less  than  P  may  cause  a  lateral  deflection,  and  that 
then  the  condition  of  indifferent  equilibrium  may  depend  upon 
the  value  of  A.  This  condition  is  probably  reached  when 
the  unit  -  stress  Sc  on  the  concave  side  reaches  the  elastic 
limit  5. 

In  conclusion  it  may  be  well  to  show  that  RANKINE'S  formula 
when  properly  deduced  is  essentially  of  the  same  form  as 


ART.  62.  ON  THE  THEORY  OF  COLUMNS.  133 

EULER'S,  and  that  the  number  q  cannot  be  a  true  constant. 
For  this  purpose  consider  the  reasoning  of  Art.  56,  and,  as 
there,  take  the  total  compressive  unit-stress  Sc  on  the  concave 
side  as  equal  to  the  sum  of  the  average  unit-stress  and  the 
flexural  unit-stress,  or, 


From  the  fundamental  formula  (4)  the  value  of  .S  is, 


Now  to  express  A  in  terms  of  /,  consider  the  case  of  columns 
with  round  ends  which  deflect  into  a  sinusoid  curve,  whose 
equation  according  to  Art.  54  is, 


y  =  A  sin  —r. 


.    nx 
T 


The  second  derivative  of  y  with  respect  to  x  is, 

d*y  An*  .  nx 
-r*  —  ~w  sm  -T-. 
dx*  I  I 

For  the  middle  of  the  column  where  x  —  £/,  the  curvature 
hence  is, 

i  _  d*y  _  An11 

«-TX*-~T- 

But  the  investigation  of  Art.  33  shows  also  that  SR  =  EC. 
Hence, 

(s     £}r 
sr  _\2 Ar 

*-^Ec-       n*Ec~  ' 
The  value  of  the  total  unit-stress  on  the  concave  side  now  is, 


134  THE   COMPRESSION   OF  COLUMNS.  CH.  V. 

and  this  is  the  same  as  (10),  except  that  q  has  been  replaced  by 

A-5 

—  ;-=:  —  ,  which  is  not  a  constant  since  it  varies  with  —  r.     This 

71  JC,  A 

p 
expression  is  a  quadratic  with  reference  to  —7,  and  by  solution 

^TL 

are  found  the  two  values, 
P 


and 


the  first  of  which  corresponds  to  the  formula  for  short  blocks 
and  the  latter  to  EULER'S  formula  for  columns  with  round  ends. 

Prob.  104.  Prove  that  Ac'  =  r*  for  a  column  so  deflected  that 
there  is  no  stress  on  the  convex  side,  c'  being  the  distance  from 
that  side  to  the  neutral  axis  of  the  cross-section. 

Prob.  105.  Prove  that  an  expression  for  the  lateral  deflection 

SI*       r* 
is  A  —  —  £g  ---  ,  and  discuss  its  meaning  when  A  =  o.     Com- 

pute the  numerical  value  of  A  for  one  of  the  examples  of  Art. 
60,  and  interpret  the  result. 


ART.  63. 


THE  PHENOMENA  OF  TORSION. 


135 


Fig.  49- 


CHAPTER  VI. 
TORSION,  AND  SHAFTS   FOR  TRANSMITTING  POWER. 

ART.  63.    THE  PHENOMENA  OF  TORSION. 

Torsion  occurs  when  applied  forces  tend  to  cause  a  twisting 
of  a  body  around  an  axis.  Let  one  end  of  a  horizontal  shaft 
be  rigidly  fixed  and 
let  the  free  end  have 
a  lever  p  attached  at 
right  angles  to  its 
axis.  A  weight  P 
hung  at  the  end  of 
this  lever  will  twist 
the  shaft  so  that  fibers 
such  as  ab,  which  were 
originally  horizontal,  assume  a  spiral  form  ad  like  the  strands 
of  a  rope.  Radial  lines  such  as  cb  will  also  have  moved  through 
a  certain  angle  bed. 

Experiments  have  proved,  that  if  P  be  not  so  large  as  to 
strain  the  material  beyond  its  elastic  limit,  the  angles  bed  and 
bad  are  proportional  to  P  and  that  on  the  removal  of  the  stress 
the  lines  cd  and  ad  return  to  their  original  positions  cb  and  ab. 
The  angle  bed  is  evidently  proportional  to  the  length  of  the 
shaft,  while  bad  is  independent  of  the  length.  If  the  elastic 
limit  be  exceeded  this  proportionality  does  not  hold,  and  if  the 
twisting  be  great  enough  the  shaft  will  be  ruptured.  These 
laws  are  but  a  particular  case  of  the  general  axioms  stated 
in  Art.  3. 

The  product  Pp  is  the  moment  of  the  force  P  with  respect  to 
the  axis  of  the  shaft,  p  being  the  perpendicular  distance  from 


136  TORSION   AND   SHAFTS.  CH.  VI. 

that  axis  to  the  line  of  direction  of  P,  and  is  called  the  twisting 
moment.  Whatever  be  the  number  of  forces  acting  at  the  end 
of  the  shaft,  their  resulting  twisting  moment  may  always  be 
represented  by  a  single  product  Pp. 

A  graphical  representation  of  the  phenomena  of  torsion  may 
be  made  as  in  Fig.  I,  the  angles  of  torsion  being  taken  as 
abscissas  and  the  twisting  moments  as  ordinates.  The  curve 
is  then  a  straight  line  from  the  origin  until  the  elastic  limit  of 
the  material  is  reached,  when  a  rapid  change  occurs  and  it 
soon  becomes  nearly  parallel  to  the  axis  of  abscissas.  The 
total  angle  of  torsion,  like  the  total  ultimate  elongation,  serves 
to  compare  the  relative  ductility  of  specimens. 

Prob.  1 06.  If  a  force  of  80  pounds  at  18  inches  from  the  axis 
twists  a  shaft  60°,  what  force  will  produce  the  same  result  when 
acting  at  4  feet  from  the  axis  ? 

Prob.  107.  A  shaft  2  feet  long  is  twisted  through  an  angle  of 
7  degrees  by  a  force  of  200  pounds  acting  at  a  distance  of  6 
inches  from  the  axis.  Through  what  angle  will  a  shaft  4  feet 
long  be  twisted  by  a  force  of  500  pounds  acting  at  a  distance 
of  1 8  inches  from  the  axis? 

ART.  64.    THE  FUNDAMENTAL  FORMULA  FOR  TORSION. 

The  stresses  which  occur  between  any  two  cross-sections  of 
a  bar  under  torsion  are  similar  to  those  of  shearing,  each  sec- 
tion tending  to  shear  off  from  the  one  ad- 
jacent to  it.  When  equilibrium  obtains  the 
external  twisting  moment  is  exactly  bal- 
anced by  the  sum  of  the  moments  of  these 
resisting  internal  stresses,  or, 

Resisting  moment  =  twisting  moment. 
The  law  governing  the  distribution  of  these 
F»g-  so-  internal  stresses  is  to  be  taken  the  same  as 

in  beams,  namely,  that  they  vary  directly  as  the  distance  from 


ART.  64.   THE   FUNDAMENTAL   FORMULA   FOR   TORSION.       137 

the  axis,  provided  that  the  elastic  limit  of  the  material  be  not 
exceeded. 

If  Pbe  the  force  acting  at  a  distance/  from  the  axis  about 
which  the  twisting  takes  place,  the  value  of  the  twisting  moment 
is  Pp.  To  find  the  resisting  moment,  let  c  be  the  distance  from 
the  axis  to  the  remotest  part  of  the  cross-section  where  the 
unit-shear  is  5S.  Then  since  the  stresses  vary  as  their  distances 
from  the  axis, 

-—  =  unit-stress  at  a  unit's  distance  from  axis, 

—  =  unit-stress  at  a  distance  z  from  axis, 
c 

-  =  total  stress  on  an  elementary  area  a, 
—  —  =  moment  of  this  stress  with  respect  to  axis, 
a  *2  =  internal  resisting  moment. 

This  may  be  written  —  2asf.     But  2as?  is  the  polar  moment 

of  inertia  of  the  cross-section  with  respect  to  the  axis,  and  may 
be  denoted  by  /.     Therefore, 


which  is  the  fundamental  formula  for  torsion. 

The  analogy  of  formula  (11)  with  formula  (4)  for  the  flexure 
of  beams  will  be  noted.  Pp,  the  twisting  moment,  is  often  the 
resultant  of  several  forces,  and  might  have  been  expressed  by 
a  single  letter  like  the  M  in  (4).  By  means  of  (11)  a  shaft 
subjected  to  a  given  moment  may  be  investigated,  or  the 
proper  size  be  determined  for  a  shaft  to  resist  given  forces. 

Prob.  108.  Three  forces  of  120,  90,  and  70  pounds  act  at 
distances  of  6,  n,  and  8  inches  from  the  axis  and  at  different 


138  TORSION  AND   SHAFTS.  Cli.  VI. 

distances  from  the  end  of  a  shaft,  the  direction  of  rotation  of 
the  second  force  being  opposite  to  that  of  the  others.  Find 
the  three  values  of  the  twisting  moment  Pp. 

Prob.  109.  A  circular  shaft  is  subjected  to  a  maximum 
shearing  unit-stress  of  2  ooo  pounds  when  twisted  by  a  force  of 
90  pounds  at  a  distance  of  27  inches  from  the  center.  What 
unit-stress  will  be  produced  in  the  same  shaft  by  two  forces  of 
40  pounds,  one  acting  at  21  and  the  other  at  36  inches  from 
the  center? 

ART.  65.    POLAR  MOMENTS  OF  INERTIA. 

The  polar  moment  of  inertia  for  simple  figures  is  readily 
found  by  the  help  of  the  calculus,  as  explained  in  works  on 
elementary  mechanics.  It  is  also  a  fundamental  principle 
that, 

• 


where  /  is  the  polar  moment  of  inertia,  7,  the  least  and  72  the 
greatest  rectangular  moment  of  inertia  about  two  axes  passing 
through  the  center.  The  following  are  values  of  J  for  some 
of  the  most  common  cases. 

For  a  circle  with  a  diameter  d,  J  —  -  , 

For  a  square  whose  side  is  d,  J  •=.  -?-, 

For  a  rectangle  with  sides  b  and  d,       J  =  --  -j  --  . 

The  value  of  c  in  all  cases  is  the  distance  from  the  axis 
about  which  the  twisting  occurs,  usually  the  center  of  figure 
of  the  cross-section,  to  the  remotest  part  of  the  cross-section. 
Thus, 

For  a  circle  with  diameter  d,  c  —  ^d, 

For  a  square  whose  side  is  d,  c  =  d  -v/f, 

For  a  rectangle  with  sides  b  and  d,     c  =  %  -\f  b*  -f-  d^. 


ART.  66.  THE  CONSTANTS  OF  TORSION.  139 

It  is  rare  in  practice  that  formulas  for  torsion  are  needed  for 
any  cross-sections  except  squares  and  circles. 

Prob.  1 10.  Find  the  values  of  J  and  c  for  an  equilateral  tri- 
angle whose  side  is  d. 

Prob.  in.  Find,  from  the  data  in  Art.  30,  the  values  of  J 
and  c  for  a  light  6  inch  /  section. 

ART.  66.    THE  CONSTANTS  OF  TORSION. 

The  constant  5,  computed  from  experiments  on  the  rupture 
of  shafts  by  means  of  formula  (11)  may  be  called  the  modulus 
of  torsion,  in  analogy  with  the  modulus  of  rupture  as  com 
puted  from  (4).  The  values  thus  found  agree  closely  with  the 
ultimate  shearing  unit-stress  given  in  Art.  7,  viz., 

For  timber,  Ss  =    2  ooo  pounds  per  square  inch, 

For  cast  iron,          .Ss  =  25  ooo  pounds  per  square  inch, 

For  wrought  iron,  5S  =  50  ooo  pounds  per  square  inch, 

For  steel,  J>s  =  75  ooo  pounds  per  square  inch. 

By  the  use  of  these  average  values  it  is  hence  easy  to  compute 

from  (n)  the  load  P  acting  at  the  distance  /  which  will  cause 

the  rupture  of  a  given  shaft. 

The  coefficient  of  elasticity  for  shearing  may  be  computed 
from  experiments  on  torsion  in  the  following  manner.  Let  a 
circular  shaft  whose  length  is  /  and  diameter  d  be  twisted 
through  an  arc  6  by  the  twisting  moment  Pp.  Here  a  point 
on  the  circumference  of  one  end  is  twisted  relative  to  a  corre- 
sponding point  on  the  other  end  through  the  arc  B  or  through 
the  distance  %6d,  so  that  the  detrusion  per  unit  of  length  is 

_w 

I'- 

From  the  fundamental  definition  of  the  coefficient  of  elasticity 
E  as  given  in  (2), 

p  _  £  _  2Ssl 
s~   8d' 


140  TORSION   AND    SHAFTS.  CH.  VI. 

and  inserting  for  Ss  its  value  from  (11),  there  results, 


from  which  E  can  be  computed  when  all  the  quantities  in  the 
second  member  have  been  determined  by  experiment,  pro- 
vided that  the  elastic  limit  of  the  material  be  not  exceeded. 
The  numerical  value  of  0  must  here  be  expressed  in  terms  of 
the  same  unit  as  n. 

Prob.  112.  What  force  P  acting  at  the  end  of  a  lever  24 
inches  long  will  twist  asunder  a  steel  shaft  1.4  inches  in 
diameter? 

Prob.  113.  An  iron  shaft  5  feet  long  and  2  inches  in  diam- 
eter is  twisted  through  an  angle  of  7  degrees  by  a  force  of 
5  ooo  pounds  acting  at  6  inches  from  the  center,  and  on  the  re- 
moval of  the  force  springs  back  to  its  original  position.  Find 
the  value  of  E  for  shearing. 

ART.  67.    SHAFTS  FOR  THE  TRANSMISSION  OF  POWER. 

Work  is  the  product  of  a  resistance  by  the  distance  through 
which  it  acts,  and  is  usually  measured  in  foot-pounds.  A  horse- 
power is  33000  foot-pounds  of  work  done  in  one  minute.  It 
is  required  to  determine  the  relation  between  the  horse-power 
H  transmitted  by  a  shaft  and  the  greatest  internal  shearing 
unit-stress  5"s  produced  in  it. 

Let  a  shaft  making  n  revolutions  per  minute  transmit  H 
horse-power  .The  work  may  be  applied  by  a  belt  from  the  motor 
to  a  pulley  on  the  shaft,  then,  by  virtue  of  the  elasticity  and  re- 
sistance of  the  material  of  the  shaft,  it  is  carried  through  other 
pulleys  and  belts  to  the  working  machines.  In  doing  this  the 
shaft  is  strained  and  twisted,  and  evidently  Ss  increases  with  H. 
Let  P  be  the  resistance  acting  at  the  circumference  of  the  pulley 
and/  the  radius  of  the  pulley.  In  making  one  revolution  the 


ART.  68.  ROUND  SHAFTS.  141 

force  P  acts  through  the  distance  2np  and  performs  the  work 
27tpP,  and  in  n  revolutions  it  performs  the  work  27tpPn.  Then 
if  P  be  in  pounds  and  p  in  inches,  the  imparted  horse-power  is, 

H_       2npPn 

33  ooo  X  12 

The  twisting  moment  Pp  in  this  expression  may  be  expressed, 
as  in  formula  (11),  by  the  resisting  moment  -^—  .  Hence  the 
equation  becomes, 

(12)  H=J^L. 

198  oooc 

This  is  the  formula  for  the  discussion  of  shafts  for  the  trans- 
mission of  power,  and  in  it  J  and  c  must  be  taken  in  inches  and 
5S  in  pounds  per  square  inch,  while  n  is  the  number  of  revolu- 
tions per  minute. 

Prob.  114.  A  wooden  shaft  6  inches  square  breaks  when 
making  40  revolutions  per  minute.  Find  the  horse-power  then 
probably  transmitted. 

ART.  68.    ROUND  SHAFTS. 

For  round  shafts  of  diameter  d,  the  values  of  /and  c  are  to 
be  taken  from  Art.  65  and  inserted  in  the  last  equation,  giving, 

or        *=  6 


The  first  of  these  may  be  used  for  investigating  the  strength 
of  a  given  shaft  when  transmitting  a  certain  number  of  horse- 
power with  a  known  velocity.  The  computed  values  of  vSs, 
compared  with  the  ultimate  values  in  Art.  67,  will  indicate  the 
degree  of  security  of  the  shaft.  Here  d  must  be  taken  in 
inches  and  Ss  will  be  in  pounds  per  square  inch. 

The  second  equation  may  be  used  for  determining  the  di- 
ameter of  a  shaft  to  transmit  a  given  horse-power  with  a  given 


142  TORSION   AND   SHAJTS.  CH.  VI. 

number  of  revolutions  per  minute.  Here  a  safe  allowable 
value  must  be  assumed  for  S,  in  pounds  per  square  inch,  and 
then  d  will  be  found  in  inches.  This  equation  shows  that  the 
diameter  of  a  shaft  varies  directly  as  the  cube  root  of  the 
transmitted  horse-power  and  inversely  as  the  cube  root  of  its 
velocity. 

Prob.  115.  Find  the  factor  of  safety  for  a  wrought  iron  shaft 
2\  inches  in  diameter  when  transmitting  25  horse-power  while 
making  100  revolutions  per  minute. 

Prob.  116.  Find  the  diameter  of  a  wrought  iron  shaft  to 
transmit  90  horse-power  with  a  factor  of  safety  of  8  when  mak- 
ing 250  revolutions  per  minute,  and  also  when  making  100 
revolutions  per  minute. 

ART.  69.    SQUARE  SHAFTS. 

For  a  square  shaft  whose  side  is  d  the  values  of  J  and  c  are 
to  be  taken  from  Art.  65  and  inserted  in  (12),  giving, 

S,  =  367500^,        or        rf 

These  are  the  same  as  for  round  shafts  except  in  the  numerical 
constants,  and  are  to  be  used  in  the  same  manner,  the  first  to 
investigate  an  existing  shaft  and  the  second  to  find  the  diameter 
for  one  proposed. 

When  a  shaft  is  used  under  practical  conditions  for  the  trans- 
mission of  power  it  usually  happens  that  it  is  subject  to  flexural 
stresses  due  to  loads  as  well  as  to  stresses  of  torsion.  Thus  a 
square  shaft  carrying  a  water  wheel  acts  like  a  beam  under  the 
weight  of  the  wheel  and  also  is  subject  to  the  twisting  moment. 
As  the  above  formulas  only  include  the  effect  of  the  latter  it  is 
clear,  if  the  transverse  load  is  considerable,  that  the  unit-stress 
.Sj  should  be  taken  small  so  as  to  allow  an  ample  margin  of 
security.  In  Art.  76  the  effect  of  transverse  loads  on  a  shaft 
will  be  taken  into  account. 


ART.  70.  MISCELLANEOUS   EXERCISES.  143 

Prob.  117.  Find  the  factor  of  safety  of  a  wooden  shaft  12 
inches  square  when  transmitting  16  horse-power  at  40  revolu- 
tions per  minute  ;  also  when  making  10  revolutions  per  minute. 

Prob.  1 1 8.  Find  the  size  of  a  square  wooden  shaft  for  a  water 
wheel  which  is  to  transmit  8  horse-power  at  20  revolutions  per 
minute  with  a  factor  of  safety  of  15. 

ART.  70.    MISCELLANEOUS  EXERCISES. 

Exercise  8.  Make  experiments  to  verify  the  phenomena  of 
torsion  stated  in  Art.  63.  Show  by  your  experiments  that  the 
strength  of  a  round  shaft  varies  directly  as  the  cube  of  its 
diameter,  and  is  independent  of  its  length. 

Exercise  9.  Make  a  theoretical  investigation  to  ascertain  if 
the  strength  of  a  square  shaft  can  be  increased  by  cutting  off 
material  from  the  corners.  If  such  is  found  to  be  the  case 
write  an  essay  explaining  the  reasoning,  the  computations  and 
the  conclusion. 

Exercise  10.  Go  to  a  testing  room  and  inspect  THURSTON'S 
testing  machine  for  torsion.  Ascertain  the  dimensions  and 
kind  of  specimens  tested  thereon.  Explain  with  sketches  the 
construction  of  the  machine,  the  method  of  its  use,  and  the 
torsion  diagrams.  State  how  the  quality  of  the  specimens  is 
inferred  from  the  torsion  diagrams. 

Prob.  1 19.  Compare  the  strength  of  a  square  shaft  with  that 
of  a  circular  shaft  of  equal  area. 

Prob.  1 20.  Jones  &  Laughlins  give,  for  computing  the  diam- 
eters of  shafts,  the  formulas, 


.  3/62.5/i 

==V"v 


and 


n  y       n 

the  first  for  ordinary  turned  wrought  iron  shafts,  and  the  sec- 
ond for  cold  rolled  wrought  iron  shafts.  What  working  unit- 
stresses  do  these  imply  ? 


144  COMBINED   STRESSES.  CH.  VII. 


CHAPTER  VII. 
COMBINED  STRESSES. 

ART.  71.  CASES  OF  COMBINED  STRESSES. 
The  three  kinds  of  simple  stress  are  tension,  compression, 
and  shear,  or,  in  other  words,  the  numerical  investigation  of 
bodies  under  stress  includes  only  the  unit-stresses  St ,  Sc ,  and  5S . 
Transverse  or  flexural  stress  was  investigated  in  Chapter  III 
by  resolving  the  internal  stresses  into  tension,  compression,  and 
shear.  Torsional  stress  is  merely  a  particular  case  of  shear. 

Tension  and  compression  are  similar  in  character  and  differ 
only  in  sign  or  direction.  Hence  their  combination  is  effected 
by  algebraic  addition.  Thus  if  P  be  a  tensile  stress  and  F  a 
compressive  stress  applied  to  the  same  bar  at  the  same  time 
the  resultant  stress  is  P—P'  which  may  be  either  tensile  or 
compressive. 

Tension  and  shear,  or  compression  and  shear,  are  often  com- 
bined, as  internally  in  the  case  of  beams  and  externally  under 
many  circumstances, 

Tension  and  flexure  are  combined  when  loads  are  placed  upon 
a  bar  under  tension.  This  case  and  that  of  compression  and 
flexure  are  of  frequent  occurrence,  and  their  investigation  is  of 
much  practical  importance. 

Flexure  and  torsion  are  combined  whenever  horizontal  shafts 
for  the  transmission  of  power  are  loaded  with  pulleys  and  belts, 
and,  as  will  be  seen,  the  effect  of  the  flexure  is  sensibly  to 
modify  the  formulas  of  the  last  chapter.  Compression,  flexure, 
and  torsion  may  occur  in  the  case  of  vertical  shafts. 


ART.  72.  STRESSES   DUE   TO   TEMPERATURE.  14$ 

The  internal  stresses  in  a  body  produced  by  applied  forces 
are  usually  of  a  complex  character.  Even  in  a  case  of  simple 
tension  there  are  shearing  stresses  in  all  directions  except  those 
perpendicular  and  parallel  to  the  line  of  tension,  as  was  shown 
in  Art.  7.  But  whatever  be  the  nature  of  the  internal  stresses 
they  may  be  investigated  by  resolving  them  into  tension,  com- 
pression, and  shear. 

Prob.  121.  A  pulls  30  pounds  at  one  end  of  a  rope,  and  B, 
C  and  D  each  pull  10  pounds  at  the  other  end.  What  is  the 
tensile  stress  in  the  rope  ? 

ART.  72.    STRESSES  DUE  TO  TEMPERATURE. 

If  a  bar  be  unstrained  it  expands  when  the  temperature  rises 
and  contracts  when  the  temperature  falls.  But  if  the  bar  be 
under  stress,  so  that  the  change  of  length  cannot  occur,  an  ad- 
ditional unit-stress  must  be  produced  which  will  be  equivalent 
to  the  unit- stress  that  would  cause  the  same  change  of  length 
in  the  unstrained  bar.  Thus  if  a  rise  of  temperature  elongates 
a  bar  of  length  unity  the  amount  s  when  free  from  stress,  it 
will  cause  the  unit-stress  S  =  sE  (see  Art.  4)  when  the  bar  is 
prevented  from  expanding  by  external  forces. 

Let  /  be  the  length  of  the  bar,  a  its  coefficient  of  linear  ex- 
pansion for  a  change  of  one  degree,  and  A  the  change  of  length 
due  to  the  rise  or  fall  of  t  degrees.  Then, 

X  =  atl. 
and  the  unit-deformation  s  is, 

A 
s  =  —.  =  at. 

The  unit-stress  produced  by  the  change  in  temperature  hence  is, 

5  =  atE 

which  is  seen  to  be  independent  of  the  length  of  the  bar.     The 
total  stress  on  the  bar  is  then  AS. 


146  COMBINED   STRESSES.  CH.  VII. 

The  following  are  average  values  of  the  coefficients  of  linear 
expansion  for  a  change  in  temperature  of  one  degree  Fahren- 
heit. 

For  brick  and  stone,  a  =  O.OOO  oo  50, 

For  cast  iron,  a  =  o.ooo  oo  62, 

For  wrought  iron,  a  =  o.ooo  oo  67, 

For  steel,  a  =  o.ooo  oo  65. 

As  an  example  consider  a  wrought  iron  tie  rod  20  feet  in 
length  and  2  inches  in  diameter  which  is  screwed  up  to  a  ten- 
sion of  9  ooo  pounds  in  order  to  tie  together  two  walls  of  a 
building.  Let  it  be  required  to  find  the  stress  in  the  rod  when 
the  temperature  falls  10°  F.  Here, 

5  =  o.ooooo  67  X  10  X  25  ooo  ooo  =  i  675  pounds. 

The  total  tension  in  the  rod  now  is, 

9  ooo  -f  3.14  X  i  675  =  14  ooo  pounds. 

Should  the  temperature  rise  10°  the  tension  in  the  rod  would 
become, 

9  ooo  —  3.14  X  i  675  =  4000  pounds. 

In  all  cases  the  stresses  caused  by  temperature  are  added  or 
subtracted  to  the  tensile  or  compressive  stresses  already 
existing. 

Prob.  122.  A  cast  iron  bar  is  confined  between  two  immovable 
walls.  What  urtit-stress  will  be  produced  by  a  rise  of  40°  in 
temperature  ? 

ART.  73.    COMBINED  TENSION  AND  FLEXURE. 

Consider  a  beam  in  which  the  flexure  produces  a  unit-stress 
5  at  the  fiber  on  the  tensile  side  most  remote  from  the  neutral 
axis.  Let  a  tensile  stress  P  be  then  applied  to  the  ends  of  the 
bar  uniformly  distributed  over  the  cross-section  A.  The  ten- 

p 
sile  unit-stress  at  the  neutral  surface  is  then  -—  and  all  the 

A 

longitudinal  stresses  due  to  the  flexure  are  increased  by  this 


ART.  73.          COMBINED  TENSION  AND  FLEXURE.  147 

p 

amount.     The  maximum  tensile  unit-stress  is  then  —  -\-  S  in 

A 
which  5  is  to  be  found  from  formula  (4). 

In  designing  a  beam  under  combined  tension  and  flexure 

p 
the  dimensions  must  be  so  chosen  that  —  -f-  5  shall  not  exceed 

A 
the  proper  allowable  working  unit-stress.  For  instance,  let  it 

be  required  to  find  the  size  of  a  square  wooden  beam  of  12 
feet  span  to  hold  a  load  of  300  pounds  at  the  middle  while 
under  a  longitudinal  stress  of  2  ooo  pounds,  so  that  the  maxi- 
mum tensile  unit-stress  may  be  about  I  ooo  pounds  per  square 
inch.  Let  d  be  the  side  of  the  square.  From  formula  (4), 
c  _  6M  _  6  X  150  X  72 

=  ^'  ~         ~d^~ 
Then  from  the  conditions  of  the  problem, 

2  ooo  .  64  800 

___  +  __.  =  I  ooo, 

from  which  results  the  cubic  equation, 

d*  —  2^=64.8, 
whose  solution  gives  for  d  the  value  4.25  inches. 

In  investigating  a  beam  under  combined  tension  and  flexure 

p 
the  maximum  value  of  —  -f-  5  is  to  be  computed,  and  the 

A 

factor  of  safety  found  by  comparing  it  with  the  ultimate  ten- 
sile strength  of  the  material. 

Prob.  123.  A  heavy  1 2-inch  I  beam  of  6  feet  span  carries  a 
uniform  load  of  200  pounds  per  linear  foot,  besides  its  own 
weight,  and  is  subjected  to  a  longitudinal  tension  of  80000 
pounds.  Find  the  factor  of  safety  of  the  beam. 

Prob.  124.  What  I  beam  of  12  feet  span  is  required  to  carry 
a  uniform  load  of  200  pounds  per  linear  foot  when  subjected 
to  a  tension  of  50  ooo  pounds,  the  maximum  tensile  stress  at 
the  dangerous  section  to  be  9  ooo  pounds  per  square  inch  ? 


148  COMBINED   STRESSES.  CH.  VII. 

ART.  74.    COMBINED  COMPRESSION  AND  FLEXURE. 

Consider  a  beam  in  which  the  flexure  produces  a  unit-stress 
5  in  the  fiber  on  the  compressive  side  most  remote  from  the 
neutral  axis.  Let  a  compressive  stress  P  be  applied  in  the  direc- 
tion of  its  length  uniformly  over  the  cross-section  A.  Then  at 

p 
the  neutral  surface  the  unit-stress  is    -  and  at  the  remotest 

A 
p 

fiber  it  is  —  -}-  S.     The  discussion  of  this  case  is  hence  exactly 
yi 

similar  to  that  of  the  last  article.  If  the  beam  is  short  the 
total  working  unit-stress  is  to  be  taken  as  for  a  short  prism  ; 
if  long  it  should  be  derived  from  RANKINE'S  formula  for 
columns. 

The  method  of  investigation  explained  in  this  and  the  pre- 
ceding article  is  the  one  ordinarily  used  in  practice  on  account 
of  the  complexity  of  the  formulas  which  result  from  the  strict 
mathematical  determination  of  the  moments  of  the  applied 
forces.  Although  not  exact  the  method  closely  approximates 
to  the  truth,  giving  values  of  the  stresses  a  little  too  large  for 
the  case  of  tension  and  a  little  too  small  for  the  case  of 
compression. 

A  rafter  of  a  roof  is  a  case  of  combined  compression  and 

flexure.  Let  b  be  its  width, 
d  its  depth,  /  the  length, 
w  the  load  per  linear  unit, 
and  0  the  angle  of  inclina- 
tion. To  find  the  horizon- 
tal reaction  H  the  center 
of  moments  is  to  be  taken 
Flg<  5I<  at  the  lower  end,  and 

,,       ,     .        ,  ,     I  COS  0  ,  „         Wl 

H  .  /  sin  0  =  ivl . ,     whence     H  =  —  cot  0. 


ART.  74.     COMBINED   COMPRESSION  AND   FLEXURE.  149 

For  any  section  whose  distance  from  the  upper  end  is  x,  the 
flexural  unit-stress  now  is  from  (4), 

-.  _  6M  _  6(Hx  sin  0  —  \wx*  cos  0) 

W*  ~  bd* 

and  the  uniform  compressive  unit-stress  is, 
c  _  H  cos  0  -f-  wx  sin  0 
S<~  ~bd~ 

The  total  compressive  unit-stress  on  the  upper  fiber  hence  is, 

c         c    I    c       3^  cos  0  (,  ,.    ,   wtcot  0  cos0      w*  sin  0 

&=£+£=  —  ^—  (/*  ___-+__ 

This  can  be  shown  to  be  a  maximum  when 


and  substituting  this,  the  maximum  unit-stress  is, 

_  32f/a  cos  0      w/cosec  0      wsin0tan0 

4^*  "   2^  ~~i~2T~ 

which  formula  may  be  used  to  investigate  or  to  design  rafters 
subject  to  uniform  loads. 

In  any  inclined  rafter  let  P  denote  all  the  load  above  a  sec- 
tion distant  x  from  the  upper  end.     Then  reasoning  as  before 
the  greatest  unit-stress  for  that  section  is  found  to  be, 
c  _  Me      P  sin  0      H  cos  0 
S*~T      ~~A~        ~A—' 
from  which  S*  may  be  computed  for  any  given  case. 

Prob.  125.  A  roof  with  two  equal  rafters  is  40  feet  in  span 
and  15  feet  in  height.  The  wooden  rafters  are  4  inches  wide 
and  each  carries  a  load  of  450  pounds  at  the  center.  Find 
the  depth  of  the  rafter  so  that  Sm  may  be  700  pounds  pet 
square  inch. 

Prob.  126.  A  wooden  beam  10  inches  wide  and  8  feet  long 
carries  a  uniform  load  of  500  pounds  per  linear  foot  and  is  sub- 
jected to  a  longitudinal  compression  of  40000  pounds.  Find 
the  depth  of  the  beam  so  that  the  maximum  working  unit-stress 
may  be  about  800  pounds  per  square  inch. 


150  COMBINED  STRESSES.  CH.  VII. 

ART.  75.  SHEAR  COMBINED  WITH  TENSION  OR  COMPRESSION. 
Let  a  bar  whose  cross-section  is  A  be  subjected  to  the  longi- 
tudinal tension  or  compression  P  and  at  the  same  time  to  a 
shear  fat  right  angles  to  its  length.     The  longitudinal  unit- 

p 

stress  is  —  which  may  be  denoted  by/,  and  the  shearing  unit- 
A 

stress  is  -  which  may  be  denoted  by  v.     It  is  required  to  find 
A 

the  maximum  unit-stresses  produced  by  the  combination  of  / 
and  v.  In  the  following  demonstration  Pwill  be  regarded  as 
a  tensile  force,  although  the  reasoning  and  conclusions  apply 
equally  well  when  it  is  compressive. 

Consider  an  elementary  cujpic  particle  with  edges  one  unit 
in  length  acted  upon  by  the  horizontal  tensile  force  /  and  /, 
and   by   the   vertical    shear  v  and  v,  as 
shown  in  Fig.  52.     These  forces  are  not  in 
equilibrium  unless  a  horizontal  couple  be 
applied  as  in  the  figure,  each  of  whose 
forces  is  equal  to  v.     Therefore  at  every 
Fig.  s*.  point  of  a  body  under  vertical  shear  there 

exists  ,a  horizontal  shear,  and  the  horizontal  shearing  unit- 
stress  is  equal  to  the  vertical  shearing  unit-stress. 

Let  a  parallelopipedal  element  have  the  length  dm,  the 
height  dn,  and  a  width  of  unity.  The  tensile  ioro.zp.dn  tends 
to  pull  it  apart  longitudinally. 
The  vertical  shear  vdn  tends  to  *\  "*' 

cause  rotation  and  this  is  resisted,       »•<?» 
as  shown  above,  by  the  horizon-  ^5re"*- 


tal   shear   vdm.      These    forces         ^^~  -^-"v.dm 

may  be  resolved  into  rectangular  v.dn\  **•<& 

components  parallel  and  perpen-  Fl*'  53' 

dicular  to  the  diagonal  ds,  as  shown  in  Fig.  53.     The  compo- 
nents parallel  to  the  diagonal  form  a  shearing  force  sdz,  and 


ART.  75.  SHEAR  COMBINED   WITH   TENSION.  151 

those  perpendicular  to  it  a  tensile  force  tdz,  s  being  the  shear- 
ing and  t  the  tensile  unit-stresses.  Let  0  be  the  angle  between 
dz  and  dm.  The  problem  is  first  to  state  expressions  for  sdz 
and  tdz  in  terms  of  0,  and  then  to  determine  the  value  of  0, 
or  the  ratio  of  dm  to  dn,  which  gives  the  maximum  values 
of  s  and  t. 

By  simple  resolution  of  forces, 

sdz  =  pdn  cos  0  -|-  vdm  cos  0  —  vdn  sin  0, 
tdz  =  pdn  sin  0  -j-  W?#  sin  0  -\-  vdn  cos  0. 

Divide  each  of  these  by  dz,  for  -3—  put  its  value  sin  0  and  for 

-T-  its  value  cos  0.     Then  the  equations  take  the  form, 

s  =  p  sin  0  cos  0  -f-  ^(cos2  0  —  sin"  0), 
t  —  p  sin8  0  -j-  2z/  sin  0  cos  0. 
These  may  be  written, 

s  =  \p  sin  20  +  v  cos  20, 

?  =  %p(i  —  cos  20)  -j-  v  sin  20. 

By  placing  the  first  derivative  of  each  of  these  equal  to  zero  it 
is  found  that, 

P 
s  is  a  maximum  when  tan  20  =  —  , 

2V 

2V 

t  is  a  maximum  when  tan  20  =  --  . 

P 

Expressing  sin  20  and  cos  20  in  terms  of  tan  20  and  inserting 
them  in  the  above  the  following  values  result  : 


max.  /  = 

These  formulas  apply  to  the  discussion  of  the  internal  stresses 
in  beams,  as  well  as  to  combined  longitudinal  stress  and  vertical 
shear  directly  applied  by  external  forces.  If  p  is  tension  t  is 


152  COMBINED   STRESSES.  CH.  VII. 

tension,  if  p  is  compression  t  is  also  compression.  If  when/  is 
tension  the  negative  sign  be  used  before  the  radical,  the  re- 
sultant value  of  t  is  the  maximum  compressive  unit-stress. 

Prob.  127.  A  bolt  f-inch  in  diameter  is  subjected  to  a  tension 
of  2  ooo  pounds  and  at  the  same  time  to  a  cross  shear  of  3  ooo 
pounds.  Find  the  maximum  tensile  and  shearing  unit-stresses 
and  the  directions  they  make  with  the  axis  of  the  bolt. 


ART.  76.    COMBINED  FLEXURE  AND  TORSION. 

This  case  occurs  when  a  shaft  for  the  transmission  of  power 
is  loaded  with  weights.  Let  5  be  the  greatest  flexural  unit- 
stress  computed  from  (4)  and  S,  the  torsional  shearing  unit- 
stress  computed^from  (12)  or  by  the  special  equations  of  Arts. 
67  and  68.  Then,  according  to  the  last  article,  the  resultant 
maximum  unit-stresses  are, 

max.  ten.  or  comp.  /  = 
max.  shear  s  =  ± 

For  wrought  iron  or  steel  it  is  usually  necessary  to  regard  only 
the  first  of  these  unit-stresses,  but  for  timber  the  second  should 
also  be  kept  in  view. 

For  example,  let  it  be  required  to  find  the  factor  of  safety  of 
a  wrought  iron  shaft  3  inches  in  diameter  and  12  feet  between 
bearings,  which  transmits  40  horse-power  while  making  120 
revolutions  per  minute,  and  upon  which  a  load  of.  800  pounds 
is  brought  by  a  belt  and  pulley  at  the  middle.  Taking  the 
shaft  as  fixed  over  the  bearings  the  flexural  unit-stress  is, 

S=  — -7,  =  5  400  pounds  per  square  inch. 
Ttd 

From  Art.  68  the  torsional  unit-stress  is, 

TT 

Ss  =  321  ooo  --T-3  =  4  ooo  pounds  per  square  inch. 


ART.  76.          COMBINED   FLEXURE  AND   TORSION.  153 

The  maximum  tensile  and  compressive  unit-stress  now  is, 

t  —  2  700  -f  -/4  ooo2  -f  2  700*  =  7  600  pounds  per  square  in. 
and  the  factor  of  safety  is  hence  over  7. 

As  a  second  example,  let  it  be  required  to  find  the  size  of 
a  square  wooden  shaft  for  a  water-wheel  weighing  3  ooo  pounds 
which  transmits  8  horse-power  while  making  20  revolutions  per 
minute.  The  length  of  the  shaft  is  16  feet,  and  one-third  of 
the  weight  is  concentrated  at  the  center  and  the  remainder  is 
equally  divided  between  two  points,  each  6  feet  from  the  center. 
Here  the  greatest  flexural  unit-stress  is, 

_  6(1  500  X  96  -  i  ooo  X  72}  _  432000 
~^~  ~~d^~' 

and  from  Art.  69  the  torsional  unit-stress  is, 

„  _  267  500  X  8  _  107000 
^~         2oJ3  d*   "' 

From  the  formula  of  the  last  Article  the  combined  tensile  or 
compressive  unit-stress  is, 

_  470400 
~~d*~' 

Now  if  the  working  value  of  t  be  taken  at  600  pounds  per 
square  inch  the  value  of  d  will  be  about  9  inches.  From  for- 
mula (13)  also 

_  254400 
d*   "' 

and  if  the  working  value  of  s  be  taken  at  150,  the  value  of  d  is 
found  to  be  about  12  inches.  The  latter  value  should  hence 
be  chosen  for  the  size  of  the  shaft. 

By  similar  reasoning  it  may  be  proved  that  the  formula  for 
finding  the  diameter  of  a  round  iron  shaft  is, 


i6M  ,   16       M*      402  500  ooo//8 

»    =  7~  +  TA  /  — f  H 2 

nt     ^     t  \]     n1  n* 


154  COMBINED   STRESSES.  CH.  VII. 

where  M  is  the  maximum  bending  moment  of  the  transverse 
forces  in  pound-inches,  H  the  number  of  transmitted  horse- 
power, n  the  number  of  revolutions  per  minute,  and  /  the 
safe  allowable  tensile  or  compressive  working  strength  of  the 
material. 

Prob.  128.  Find  the  factor  of  safety  for  the  data  of  Prob. 
115  when  the  shaft  is  in  bearings  12  feet  apart  and  carries  a 
load  of  200  pounds  at  the  middle. 

ART.  77.    COMBINED  COMPRESSION  AND  TORSION. 

In  the  case  of  a  vertical  shaft  the  torsional  unit-stress  Sj  com- 
bines with  the  direct  compressive  stress  due  to  the  weights 
upon  the  shaft,  and  produces  a  resultant  compression  t  and 
shear  s.  From  formulas  (13)  the  combined  unit-stresses  are, 


The  use  of  these  is  the  same  as  those  of  the  last  Article,  Ss 
being  found  from  the  formulas  of  Chapter  VI,  while  Sc  is  com- 
puted from  formula  (i)  if  the  length  of  the  shaft  be  less  than 
ten  times  its  diameter  and  from  (10)  for  greater  lengths. 

In  order  to  prevent  vibration  and  flexure  it  is  usual  to  place 
bearings  at  frequent  intervals  on  a  vertical  shaft  so  that  prob- 
ably the  use  of  formula  (10)  will  rarely  be  required,  particularly 
if  /  be  taken  at  a  low  value.  For  a  round  shaft  the  expression 
for  t  becomes, 

_  4/>  I  T777      i6/^ 

~~  Ttd*  ~*~  \  ri*d*      7t*d* ' 

in  which  P  is  the  load.  From  this  the  diameter  d  may  be  found 
when  t  and  the  other  data  are  given. 

Prob.   129.  A  vertical  shaft,  weighing  with   its  loads   6000 


ART.  78.  HORIZONTAL  SHEAR  IN  BEAMS.  155 

pounds,  is  subjected  to  a  twisting  moment  by  a  force  of  300 
pounds  acting  at  a  distance  of  4  feet  from  its  center.  If  the 
shaft  is  wrought  iron,  4  feet  long  and  2  inches  in  diameter,  find 
its  factor  of  safety. 

Prob.  130.  Find  the  diameter  of  a  short  vertical  steel  shaft 
to  carry  loads  amounting  to  6  ooo  pounds  when  twisted  by  a 
force  of  300  pounds  acting  at  a  distance  of  4  feet  from  the 
center,  taking  the  unit-stress  against  compression  as  10  ooo  and 
against  shearing  as  7  ooo  pounds  per  square  inch. 

ART.  78.    HORIZONTAL  SHEAR  IN  BEAMS. 

The  common  theory  of  flexure  as  presented  in  Chapters  III 
and  IV  considers  that  the  internal  stresses  at  any  section  are 
resolved  into  their  horizontal  and  vertical  components,  the 
former  producing  longitudinal  tension  and  compression  and 
the  latter  a  transverse  shear,  and  that  these  act  independently 
of  each  other.  Formula  (3)  supposes  further  that  the  vertical 
shear  is  uniformly  distributed  over  the  cross-section  of  the 
beam.  A  closer  analysis  will  show  that  a  horizontal  shear 
exists  also  and  that  this,  together  with  the  vertical  shear, 
varies  in  intensity  from  the  neutral  surface  to  the  upper  and 
lower  sides  of  the  beam.  It  is  well  known  that  a  pile  of  boards 
which  acts  like  a  beam  deflects  more  than  a  solid  timber  of  the 
same  depth,  and  this  is  largely  due  to  the  lack  of  horizontal 
resistance  between  the  layers.  The  common  theory  of  flexure 
in  neglecting  the  horizontal  shear  generally  errs  on  the  side  of 
safety.  In  a  few  experiments  however  beams  have  been  known 
to  crack  along  the  neutral  surface  and  it  is  hence  desirable  to 
investigate  the  effect  of  horizontal  shear  in  tending  to  cause 
rupture  of  that  kind.  That  a  horizontal  shear  exists  simulta- 
neously with  the  vertical  shear  is  evident  from  the  considera- 
tions in  Art.  75. 

Let  Fig.  54  represent  a  portion  of  a  bent  beam  of  uniform 


I56 


COMBINED   STRESSES. 


CH.  VII. 


section.  Let  a  rectangular  notch  nmpq  be  imagined  to  be  cut 
into  it,  and  let  forces  be  applied  to  it  to  preserve  the  equilib- 
rium. Let  H  be  the  sum  of  all  the  horizontal  components  of 

these  forces  act- 


i%JP    j 

t 

m__ 

ing  on  mn  and  n 
m'    the  sum  of  those 

Now//7  is  greater 

Fig.  54- 

hence  the  differ- 

ence  H'  —  H  must  act  along  mq  as  a  horizontal  shear.  Let  the 
distance  mq  be  dx,  the  thickness  mm'  be  b,  and  the  area  mqmm' 
be  at  a  distance  c'  above  the  neutral  surface.  Let  c  be  the  dis- 
tance from  that  neutral  surface  to  the  remotest  fiber  where  the 
unit-stress  is  S.  Let  a  be  the  cross-section  of  any  fiber.  Let. 
M  be  the  bending  moment  at  the  section  mn  and  M'  that  at 
the  section  qp.  Now  from  the  fundamental  laws  of  flexure, 

—  =  unit-stress  at  a  unit's  distance  from  neutral  surface, 
c 

—  y  =  unit-stress  at  distance  y  from  neutral  surface, 
— —  =  total  stress  on  fiber  a  at  distance  y, 


—  sum  of  horizontal  stresses  between  m  and  «. 


o          Tif 

The  value  of  H  hence  is,  since  -==-—, 

c       I 


and  likewise  for  the  other  section, 


ART.  78.  HORIZONTAL  SHEAR  IN  BEAMS.  157 

The  horizontal  shear  therefore  is  expressed  by 


Now  since  the  distance  mq  is  dx,  the  value  of  M1  —  M  is  dM. 
Also  if  Sh  be  the  horizontal  shearing  unit-stress  upon  the  area 
bdx  the  value  of  H'  —  H  is  Shbdx.     Hence, 
dM 


Again  from  Art.  45  it  is  plain  that  —  —  is  the  vertical  shear  V 
at  the  section  under  consideration.     Therefore, 

(H)  Sk  =  -- 

lo 

is  the  formula  for  the  horizontal  shearing  unit-stress  at  any 
point  of  any  section  of  the  beam. 

This  expression  shows  that  the  horizontal  shearing  unit-stress 
is  greatest  at  the  supports,  and  zero  at  the  dangerous  section 
where  V  is  zero.  The  summation  expression  is  the  statical 
moment  of  the  area  mm'nn'  with  reference  to  the  neutral  axis  ; 
it  is  zero  when  y  =  c,  and  a  maximum  when  y  =  o.  Hence 
the  longitudinal  unit-shear  is  zero  at  the  upper  and  lower  sides 
of  the  beam  and  is  a  maximum  at  the  neutral  surface.  The 
formula  for  the  maximum  horizontal  shearing  unit-stress  at  any 
section  therefore  is, 


Here  /  is  the  moment  of  inertia  of  the  whole  cross-section  with 
reference  to  the  neutral  axis  (Art.  23),  b  is  the  width  of  the 
beam  along  the  neutral  surface,  and  "2?0ay  is  the  statical  mo- 
ment of  the  area  of  the  part  of  the  cross-section  on  one  side  of 
the  neutral  axis.  Let  Al  be  the  area  of  the  cross-section  on 


158  COMBINED   STRESSES.  CH.  VII. 

one  side  of  the  neutral  axis  and  cv  the  distance  of  its  center  of 
gravity  from  that  axis;  then  2c0ay  =  A1cl,  and  the  formula 
becomes, 

,     v  c        VAlCl 

(14)'  s*=-7r> 

which  gives  the  maximum  shearing  unit-stress,  both  horizontal 
and  vertical,  at  the  neutral  surface.  The  mean  unit-stress  given 
by  (3)  is  always  less  than  this  maximum. 

For  a  rectangular  beam  of  breadth  b  and  depth  d,  the  value 

.   bd*  bd    d      bd* 

of  /  is  -  ,  and  Alcl  =  —  .  -  =  —  —  .     Then, 


-  ~2bd' 

By  inserting  in  this  the  value  of  V  for  any  section  the  corre- 
sponding value  of  Ss  at  the  neutral  surface  is  found.  In  this 
particular  instance  it  is  seen  that  the  approximate  formula  (3) 
gives  values  of  St  which  are  33  per  cent  lower  than  the  true 
maximum  value. 

Prob.  131.  In  the  Journal  of  the  Franklin  Institute  for  Feb- 
ruary, 1883,  is  detailed  an  experiment  on  a  spruce  joist  3-f  X  12 
inches  and  14  feet  long,  which  broke  by  tension  at  the  middle 
and  afterwards  by  shearing  along  the  neutral  axis  at  the  end 
when  loaded  at  the  middle  with  12  545  pounds.  Find  the  ten- 
sile and  shearing  unit-stresses. 

ART.  79.    MAXIMUM  INTERNAL  STRESSES  IN  BEAMS. 

From  the  last  Article  it  is  evident  that  at  every  point  of  a 
beam  there  exists  a  horizontal  unit-shear  of  the  intensity  S;t  and 
also  a  vertical  unit-shear  of  the  same  intensity,  whose  value  is 
given  by  (14).  At  every  point  there  also  exists  a  longitudinal 
tension  or  compression  which  may  be  computed  from  (4)  with 
the  aid  of  the  principle  that  these  stresses  vary  directly  as  their 


ART.  79.    MAXIMUM   INTERNAL  STRESSES   IN   BEAMS.  1  59 

distances  from  the  neutral  axis.  Let  v  denote  the  unit-shear 
thus  determined  and  p  the  tensile  or  compressive  unit-stress. 
Then  from  Art.  75  the  maximum  unit-shear  at  that  point  is, 

s  =  jj  +  i/, 

and  it  makes  an  angle  0  with  the  neutral  surface  such  that, 
tan  20  =-^-. 

2V 

Also  the  maximum  tensile  or  compressive  unit-stress  at  that 
point  is, 


and  it  makes  an  angle  6  with  the  neutral  surface  such  that, 

tan  20=-  —  . 
P 

From  these  formulas  the  lines  of  direction  of  the  maximum 
stresses  may  be  traced  throughout  the  beam. 

For  the  maximum  shear  v  is  greatest  and  p  is  zero  at  the 
neutral  surface,  while  v  is  zero  and  p  is  greatest  at  the  upper 
and  lower  surfaces.  Hence  for  the  neutral  surface  0  is  o,  it 
increases  with  p,  and  becomes  45°  at  the  upper  and  lower 
surfaces. 

For  the  maximum  tension  /  is  greatest  and  equal  to  /  on 
the  convex  side  where  v  =  o  and  6  =  o.  As  the  neutral  sur- 
face is  approached  v  increases,  /  decreases,  and  6  increases. 
At  the  neutral  surface  v  is  greatest,  p  is  zero,  and  0  —  —  45°. 
Here  the  maximum  tension  and  compression  are  each  equal 
to  v. 

For  the  maximum  compression  in  like  manner  8  is  o°  at  the 
concave  surface  and  45°  at  the  neutral  surface.  The  lines  of 
maximum  tension  if  produced  beyond  the  neutral  surface  would 
evidently  cut  those  of  maximum  compression  at  right  angles 
and  be  vertical  at  the  concave  surface. 


l6o  COMBINED   STRESSES.  CH.  VII. 

The  following  figure  is  an  attempt  to  represent  the  lines 
which  indicate  the  directions  of  the  maximum-  unit-stress  in  a 

beam.  The  full  lines 
above  the  neutral 
surface  are  those  of 
maximum  compres- 
sion, while  those  be- 
low are  maximum 
tension.  The  broken 
lines  are  those  of 
Fig- 55>  maximum  shear.  On 

any  line  the  intensity  of  stress  varies  with  the  inclination,  being 
greatest  where  the  line  is  horizontal  and  least  where  its  inclina- 
tion is  45°.  The  lines  of  maximum  shear  cut  those  of  maxi- 
mum tension  and  compression  at  angles  of  45°.  The  lines  of 
maximum  tension  above  the  neutral  surface  and  those  of  maxi- 
mum compression  below  it  are  not  shown  ;  if  drawn  they  would 
cut  the  others  at  right  angles  and  become  vertical  at  the  upper 
and  lower  edges  of  the  beam. 

It  appears  from  the  investigation  that  the  common  theory  of 
flexure  gives  the  horizontal  unit-stress  correctly  at  the  dan- 
gerous section  of  a  simple  beam  where  the  vertical  shear  is 
zero.  At  other  sections  the  stress  5"  as  computed  from  (4)  is 
correct  for  the  remotest  fiber,  but  for  other  fibers  the  unit-stress 
/  is  greater.  It  is  hence  seen  that  the  main  practical  value  of 
the  theory  of  internal  stress  is  in  showing  that  the  intensity  of 
the  shear  varies  throughout  the  cross-section  of  the  beam.  For 
a  restrained  beam,  where  the  vertical  shear  suddenly  changes 
sign  at  the  dangerous  section,  the  common  theory  gives  the 
horizontal  stress  5  correctly  for  the  remotest  fiber  only,  and 
it  might  be  possible  in  some  forms  of  cross-sections  for  the 
maximum  stress  t  to  be  slightly  greater  than  6"  for  a  fiber 
nearer  to  the  neutral  surface.  All  that  has  here  been  deduced 


ART.  79.     MAXIMUM   INTERNAL  STRESSES   IN   BEAMS.  l6l 

justifies  the  validity  of  the  common  theory  of  flexure  as  a 
correct  guide  in  the  practical  design  and  investigation  of 
beams. 

Prob.  132.  A  joist  fixed  at  both  ends  is  3  X  12  inches  and 
12  feet  long,  and  is  strained  by  a  load  at  the  middle,  so  that 
the  value  of  5  as  computed  from  (4)  is  4  ooo  pounds  per  square 
inch.  Find  the  value  of  t  for  points  over  the  support  distant 
3,  4,  and  5  inches  from  the  neutral  surface. 

Prob.  133.  Show,  for  a  point  between  the  neutral  surface  and 
the  convex  side,  that  there  exists  a  maximum  compression  as 
well  as  a  maximum  tension.  Deduce  an  expression  for  the 
value  of  this  maximum  compression  and  its  direction.  Draw 
a  figure  showing  the  curves  over  the  entire  beam  for  both  these 
stresses. 


1 62  APPENDIX  AND  TABLES.  CH.  VIII. 


CHAPTER  VIII. 
APPENDIX  AND  TABLES. 

ART.  80.  SUDDEN  LOADS  AND  SHOCKS. 
When  a  tensile  load  is  gradually  applied  to  a  bar  its  intensity 
increases  slowly  from  o  up  to  the  final  value  P,  and  the  stress 
in  the  bar  at  any  instant  is  equal  to  the  tensile  force  existing 
at  that  instant  ;  the  elongation  of  the  bar  increases  propor- 
tionally to  the  stress  from  o  up  to  the  final  limit  A,  if  the 
elastic  limit  is  not  exceeded.  The  work  done  upon  the  bar  by 
the  external  force  is  then  equal  to  its  mean  intensity  \P  multi- 
plied by  the  distance  A,  or  |/>A  ;  the  work  of  the  molecular 
forces  is  also  equal  to  this  same  quantity  £/>A. 

A  load  P  is  said  to  be  suddenly  applied  when  its  intensity  is 
the  same  from  the  beginning  to  the  end  of  the  elongation. 
The  stress  in  the  bar,  however,  increases  from  o  up  to  a  limit  Q. 
Let  y  be  the  elongation  produced  by  the  sudden  load  P\  then 
the  work  of  this  external  force  is  Py.  If  the  stresses  are 
within  the  elastic  limit  so  that  they  increase  proportionally  to 
the  elongation,  the  mean  stress  is  \Q  and  the  work  of  the  re- 
sisting forces  is  \Qy.  Hence,  as  these  two  works  must  be  equal, 

\Qy  =  Py         or         Q  =  2P. 

Now  let  A  be  the  elongation  due  to  the  load  P  when  gradually 
applied,  then  by  law  (B), 


Therefore  is  established  the  following  important  theoretical  law, 
A  suddenly  applied  load  produces  double  the  stress  and 
double  the  deformation  caused  by  the  same  load  when 
gradually  applied. 


ART.  80.  SUDDEN   LOADS   AND   SHOCKS.  163 

This  law  is  only  true  when  all  the  stresses  are  within  the  elastic 
limit  of  the  material.  The  sudden  load  P  thus  causes  the  end 
of  the  bar  to  move  from  o  to  2\  when  the  stress  becomes  2.P 
the  resultant  force  tending  to  move  the  end  is  P  —  2P  or  —  P 
and  hence  the  end  moves  backward,  until  after  a  series  of 
oscillations  it  comes  to  rest  with  the  elongation  A  due  to  the 
static  stress  P.  The  time  of  this  oscillation,  as  also  the  velocity 
of  the  end  of  the  bar  at  any  instant,  can  be  computed  by  the 
principles  of  dynamics. 

A  shock  is  said  to  be  produced  upon  the  end  of  a  bar  when 
a  load  P  falls  from  a  height  h  upon  it.  Here  the  stress  in  the 
bar  will  increase  from  o  up  to  a  certain  limit  Q  and  the  defor- 
mation from  o  up  to  a  certain  limit  y.  If  the  elastic  limit  of 
the  material  be  not  exceeded,  the  stress  at  any  instant  will  be 
proportional  to  the  deformation,  so  that  the  work  of  the  in- 
ternal stresses  will  be  ^Qy,  The  work  done  by  the  exterior  force 
P  in  the  same  time  is  P(Ji  -|-  y).  Hence 


But  if  A  be  the  deformation  due  to  a  static  load  P,  the  law  of 
proportionality  gives 

Q_l 
p  ~A- 

Combining  these  two  equations  there  is  found, 


If  k  =  O  these  formulas  reduce  to  Q  =  2P  and  y  =  2A,  which 
is  the  case  of  a  suddenly  applied  load ;  if  h  =  4j,  they  become 
Q  =  4P  and  y  =  4^  ;  if  h  =  i2A  they  give  Q  =  6P  and  y  =  6A. 
Since  A  is  a  small  quantity  for  any  metallic  bar,  it  follows  that 
a  load  P  dropping  from  a  moderate  height  may  produce  great 


164  APPENDIX   AND   TABLES.  CH.  VIII. 

stresses  and  deformations.  Experiments  made  upon  springs 
show  that  the  theory  here  presented  is  correct,  provided  the 
elastic  limit  of  the  material  is  not  surpassed  by  the  stress  Q. 

The  effect  of  loads  applied  with  shock  is  therefore  to  cause 
stresses  and  deformations  greatly  exceeding  those  produced  by 
the  same  static  loads,  so  that  the  elastic  limit  may  perhaps  be 
often  exceeded.  Moreover  the  rapid  oscillations  and  the  rapid 
variations  in  the  stresses  cause  a  change  in  molecular  structure 
which  impairs  the  elasticity  of  the  material.  Generally  it  will 
be  found  that  the  appearance  of  a  fracture  of  a  bar  which  has 
been  subject  to  shocks  is  of  a  crystalline  nature,  whereas  the 
same  material,  if  ruptured  under  a  gradually  increasing  stress, 
would  exhibit  a  tough  fibrous  structure.  Shocks  which  produce 
stresses  above  the  elastic  limit  cause  the  material  to  become 
stiff  and  brittle,  and  hence  it  is  that  the  working  unit-stresses 
based  upon  static  loads  should  be  taken  very  low  (Art.  8). 

Prob.  134.  In  an  experiment  upon  a  spring  a  weight  of  14.79 
ounces  produced  an  elongation  of  0.42  inches,  but  when 
dropped  from  a  height  of  7.72  inches  it  produced  a  stress  of  102.3 
ounces  and  an  elongation  of  2.90  inches.  Compare  theory  with 
experiment. 

ART.  81.    THE  RESILIENCE  OF  MATERIALS. 

When  an  applied  stress  causes  a  deformation  work  is  done. 
Thus  if  a  tensile  stress  P  be  applied  by  increments  to  a  bar, 
so  that  the  stress  gradually  increases  from  o  to  the  value  P,  the 
work  done  is  the  product  of  the  average  stress  by  the  total 
elongation  A.  This  product  is  termed  the  resilience  of  the  bar. 
If  the  stress  does  not  exceed  the  elastic  limit  of  the  material 
the  average  stress  is  %P,  and  the  work  or  resilience  is  £/>A.  If 
the  cross-section  of  the  bar  be  A  and  its  length  /,  the  unit-stress 

is  -    or  S,  and  the  unit-strain  is      or  s,  so  that  the  work  done 


ART    8l.  THE   RESILIENCE  OF  MATERIALS.  l6$ 

on  each  unit  of  length  of  the  bar  per  unit  of  cross-section  is 
%Ss.  From  formula  (2)  the  value  of  s  is  ^,  and  accordingly  this 
work  may  be  written, 

(16)  K=1-^. 

If  6"  be  the  unit-stress  at  the  elastic  limit,  the  quantity  K  is 
called  the  modulus  of  resilience  of  the  material. 

Resilience  is  often  regarded  as  a  measure  of  the  capacity  of 
a  material  to  withstand  shock,  for  if  a  shock  or  sudden  stress 
be  produced  by  a  falling  body,  its  intensity  depends  upon  the 
weight  and  the  height  through  which  it  has  fallen,  that  is,  upon 
its  kinetic  energy  or  work.  Hence  the  higher  the  resilience  of  a 
material  the  greater  is  its  capacity  to  endure  work  that  may  be 
performed  upon  it.  The  modulus  of  resilience  is  a  measure  of 
this  capacity  within  the  elastic  limit  only. 

The  following  are  values  of  the  modulus  of  resilience  as 
computed  from  (16)  by  the  use  of  the  average  constants  given 
in  Art.  5. 

For  timber,  K—    3.0  inch-pounds, 

For  cast  iron,  K=    1.2  inch-pounds, 

For  wrought  iron,        K=  12.5  inch-pounds, 
For  steel,  K  =  41.7  inch-pounds. 

The  ultimate  resilience  of  materials  cannot  be  expressed  by  a 
rational  formula,  because  the  law  of  increase  of  elongation  be- 
yond the  elastic  limit  is  unknown.  In  Fig.  I  the  ultimate 
resilience  is  indicated  by  the  area  between  any  curve  and  the 
axis  of  abscissas,  since  that  area  has  the  same  value  as  the  total 
work  performed  in  producing  rupture.  For  timber  and  cast 
iron  the  ratio  of  these  areas  is  about  the  same  as  that  of  the 
values  of  K,  but  for  wrought  iron  and  steel  the  areas  are  nearly 
equal. 

Prob.  135.  What  horse-power  engine  is  required  to  strain  125 


l66  APPENDIX  AND   TABLES.  CM.  VIII. 

times  per  minute  a  bar  of  wrought  iron  2  inches  in  diameter 
and  1 8  feet  long,  from  o  up  to  one-half  its  elastic  limit? 

ART.  82.    THE  FATIGUE  OF  MATERIALS. 

The  ultimate  strength  Su  is  usually  understood  to  be  that 
steady  unit-stress  which  causes  rupture  at  one  application. 
Experience  and  experiments,  however,  teach  that  if  a  unit- 
stress  somewhat  less  than  Su  be  applied  a  sufficient  number  of 
times  to  a  bar  rupture  will  be  caused.  The  experiments  of 
W6HLER  have  been  of  great  value  in  establishing  the  laws 
which  govern  the  rupture  of  metals  under  repeated  applica- 
tions of  stress.  For  instance,  he  found  that  the  rupture  of  a 
bar  of  wrought  iron  by  tension  was  caused  in  the  following 
different  ways. 

By  800  applications  of  52  800  pounds  per  square  inch. 

By        107  ooo  applications  of  48  400  pounds  per  square  inch. 

By       450  OOO  applications  of  39  ooo  pounds  per  square  inch. 

By  10  140  ooo  applications  of  35  ooo  pounds  per  square  inch. 
The  range  of  stress  in  each  of  these  applications  was  from  o  to 
the  designated  number  of  pounds  per  square  inch.  Here  it  is 
seen  that  the  breaking  stress  decreases  as  the  number  of  appli- 
cations increase.  In  other  experiments  where  the  initial  stress 
was  not  o,  but  a  permanent  value  S,  the  same  law  was  seen  to 
hold  good.  It  was  further  observed  that  a  bar  could  be  strained 
from  o  up  to  a  stress  near  its  elastic  limit  an  enormous  number 
of  times  without  rupture.  From  a  discussion  of  these  numer- 
ous experiments  the  following  laws  may  be  stated. 

1.  By  repeated  applications  of  stress  rupture  may  be  caused 
by  a  unit-stress  less  in  value  than  the  ultimate  strength 
of  the  material. 

2.  The  greater  the  range  of  stress  the  less  is  the  unit-stress 
required  to  produce  rupture  after  an  enormous  number 
of  applications. 

3.  When  the  stress  ranges  from  o  up  to  a  value  about  equal 


ART.  83.  WORKING  STRENGTHS  FOR  REPEATED  STRESSES.  l6/ 

to  the  elastic  limit  the  number  of  applications  required 
to  rupture  it  is  enormous. 

4.  A  range  of  stress  from  tension  into  compression,  or  vice 
versa,  produces  rupture  with  a  less  number  tof  applica- 
tions than  the  same  range  in  stress  of  one  kind  only. 

5.  When  the  range  of  stress  in  tension  is  equal  to  that  in 
compression  the  stress  which  will  produce  rupture  after 
an  enormous  number  of  applications  is  a  little  greater 
than  one-half  the  elastic  limit. 

The  term  '  enormous  number '  used  in  stating  these  laws 
means  about  40  millions,  that  being  roughly  the  number  used 
by  WOHLER  to  cause  rupture  under  the  conditions  stated. 
For  all  practical  cases  of  repeated  stress,  except  in  fast  moving 
machinery,  this  great  number  would  seldom  be  exceeded  during 
the  natural  life  of  the  piece. 

In  Art.  8  it  was  recognized  that  the  working  strength  should 
be  less  for  pieces  subject  to  varying  stresses  than  for  those  car- 
rying steady  loads  only.  For  many  years  indeed  it  has  been 
the  practice  of  designers  to  grade  the  working  strength  accord- 
ing to  the  range  of  stresses  to  which  it  might  be  liable  to  be 
subjected.  WOHLER'S  laws  and  experiments  afford  however  a 
means  of  grading  these  values  in  a  more  satisfactory  manner 
than  mere  judgment  can  do,  and  formulas  for  that  purpose  will 
be  deduced  in  the  next  Article.  After  the  working  strength 
Sv  is  determined  the  cross-section  of  the  piece  is  found  in  the 
usual  way,  if  in  tension  by  formula  (i),  and  if  in  compression 
by  formula  (i)  or  (10)  as  the  case  may  require. 

Prob.  136.  How  many  years  will  probably  be  required  for  a 
tie  bar  in  a  bridge  truss  to  receive  40  million  repetitions  of 
stress  ? 

ART.  83.    WORKING  STRENGTHS  FOR  REPEATED  STRESSES 

Consider  a  bar  in  which  the  unit-stress  varies  from  S'  to  S, 
the  latter  being  the  greater  numerically.  Both  S'  and  S  may 


i68 


APPENDIX   AND   TABLES. 


CH.  VIII. 


be  tension  or  both  may  be  compression,  or  one  may  be  tension 
and  the  other  compression.  In  the  last  case  the  sign  of  Sf  is 
to  be  taken  as  minus.  Consider  the  stress  to  be  repeated  an 
enormous  number  of  times  and  rupture  to  then  occur.  By  the 
second  law  above  stated  5  is  some  function  of  the  range  of 
stres,s  or, 


This  may  be  expressed  in  another  way,  thus, 


or,  in  words,  the  rupturing  stress  5  after  an  enormous  number 
of  repetitions  is  a  function  of  the  ratio  of  the  limiting  stresses. 

Let  u  be  the  utimate  strength  of  the  material,  tensile  if  5  is 
tension  and  compressive  if  .S  is  compression.  Let  e  be  the 
unit-stress  at  the  elastic  limit,  and  f  the  unit-stress  which  pro- 
duces rupture  after  an  enormous  number  of  repetitions  when 
the  range  of  stress  in  tension  is  equal  to  that  in  compression. 
It  is  required  to  find  the  value  of  5  in  terms  of  u,  e,  /,  and  the 

ratio  ^r.     For  this  purpose  let  the  values  of  the  ratio  be  re- 

garded as  abscissas  and  those  of  *S  as  ordinates,  the  former 

ranging  from  -j-  i  to  —  I  as 
seen  in  the  figure.  Now  if 
this  ratio  is  -}-  I  there  is  no 
range  of  stress  and  S  =  u 
as  in  cases  of  steady  load. 
Again  when  the  ratio  is  o 
the  third  law  gives  S  =  e  • 
ancl  lastl  when  the  ratio 


o 
-  SB. 


is  —  i  the  fifth  law  gives 
S  =f.  The  most  rational  assumption  as  to  the  law  of  variation 
of  S  is  that  it  represented  by  some  curve  passing  through  the 


ART.  83.  WORKING  STRENGTHS  FOR  REPEATED  STRESSES.    169 

three  points  determined  by  the  ordinates  u,  e,  and  f.     The 
simplest  curve  is  a  parabola,  whose  equation  is, 


in  which  m,  n,  and/  are  quantities  to  be  determined  from  the 
conditions  just  stated,  and  doing  this  there  results 


This  formula  is  not  to  be  regarded  as  the  true  law  of  rupturing 
strength  under  repeated  stresses,  but  merely  as  an  empirical 
statement  which  agrees  with  the  limiting  values  determined  by 
experiment,  and  which  will  give  approximately  intermediate 
values. 

The  formulas  most  frequently  used  for  determining  the  unit- 
stress  which  will  cause  rupture  under  repeated  loads  are  those 
of  LAUNHARDT  and  WEYRAUCH,  that  of  the  former  being  ap- 
plicable when  the  limiting  stresses  S'  and  .S  are  both  tension  or 
both  compression,  and  that  of  the  latter  when  one  limiting 
stress  is  tension  and  the  other  compression.  LAUNHARDT  sup- 
poses that  S  varies  uniformly  between  the  ordinates  u  and  e  so 
that  its  equation  is  that  of  a  straight  line,  or 


and  the  graphical  representation  is  that  of  the  straight  line  in 
the  right  hand  part  of  Fig.  56.  It  is  seen  that  formula  (17) 
gives  values  of  S  slightly  less  than  those  from  LAUNHARDT'S, 
except  for  the  ratios  o  and  I  when  they  agree. 

The  formula  of  WEYRAUCH  applies  to  the  case  where  the 
range  of  stress  is  from  tension  into  compression  or  vice  versa, 
and  it  also  supposes  the  law  of  variation  to  be  that  of  a  straight 
line  between  the  limiting  ordinates  given  by  experiment,  or 


I/O  APPENDIX  AND   TABLES.  CH.  VIII. 

in  which  the  numerical  value  of  the  ratio  S'  :  S  is  to  be  taken 
as  positive.  This  equation  is  represented  by  the  straight  line 
in  the  left  hand  part  of  Fig.  56.  Here  also  formula  (17) 
gives  less  values  for  5  than  those  obtained  by  WEYRAUCH'S 
formula. 

In  designing  a  bar  which  is  to  be  'subject  to  an  enormous 
number  of  repetitions  of  stress,  ranging  from  P'  to  P,  the  ratio 

P'  S' 

—-  is  the  same  as  — ,  and  formula  (17)  gives  the  unit-stress  5 

which  will  cause  rupture  after  an  enormous  number  of  repeti- 
tions. To  be  sure  of  safety  a  factor  of  security  must  be  applied ; 
then  the  working  unit-stress  is  found  by  dividing  5  by  this  fac- 
tor, which  is  here  usually  taken  the  same  as  the  factor  of  safety 
for  a  steady  load  where  there  is  no  range  of  stress.  For  ex- 
ample, consider  a  kind  of  wrought  iron  for  which  w  =  52  ooo, 
e  =  26000,  and/"=  13  ooo  pounds  per  square  inch,  and  let  the 
factor  of  security  be  4.  Then  formula  (17)  becomes, 


from  which  the  allowable  value  cf  the  working  unit-stress  can 
be  computed  for  assigned  values  v_/f  the  ratio  S'  :  S. 

For  example,  let  it  be  required  to  find  the  proper  cross-sec- 
tion of  a  wrought  iron  bar  which  is  to  be  subjected  to  a  repeated 
tension  ranging  from  30  ooo  pounds  under  dead  load  to  90  ooo 
pounds  under  full  live  load.  Here 

S^_P'  _  30000  _  i_ 
S  ~  P  ~  90  ooo  ~~  3 ' 
and  from  the  formula  just  deduced, 

Sw  =  6$oo(i  -[-£.£  _[-£.£)  =  8300. 
Then  the  cross-section  of  the  bar  is, 

A  = =  10.9  square  inches. 


ART.  84.  THE   INTERNAL  WORK  IN  BEAMS.  I/I 


S'  _  —  30000  _   _i_ 

S         +QOOOO~         *' 


But  if  the  bar  is  to  be  subjected  to  repeated  stress  varying  from 
30  ooo  pounds  compression  to  90  ooo  pounds  tension,  then 

5'  = 

+  90000 

and  from  the  special  formula, 

S,  =  6  500(1  -  |-i  +  i-i)  =  5  050, 
so  that  the  cross-section  of  the  bar  should  be, 

A  =  — =  17.8  square  inches, 

5050 

which  is  63  per  cent  larger  than  required  for  the  smaller  range. 

The  quantity /which  is  the  unit-stress  required  to  produce 
rupture  after  an  enormous  number  of  repetitions  in  alternating 
stress  of  equal  amplitudes,  is  called  the  '  vibration  strength '  by 
some  writers.  Its  value  for  wrought  iron  is  about  one-half  and 
for  steel  a  little  greater  than  one-half  the  elastic  limit.  For 
other  materials  there  is  as  yet  no  experimental  knowledge 
regarding  its  value. 

Prob.  137.  A  steel  bar  one  inch  in  diameter  is  subject  to  re- 
peated stress  ranging  between  1 5  ooo  pounds  tension  and  40  ooo 
pounds  tension.  Will  it  break  after  an  enormous  number  of 
repetitions  ? 

Prob.  138.  Show  that,  according  to  the  above  investigation, 
the  working  unit-stress  for  wrought  iron  bars  subject  to  re- 
peated applications  of  equal  tension  and  compression  should 
be  about  one-fourth  of  that  for  a  steady  stress. 

ART.  84.    THE  INTERNAL  WORK  IN  BEAMS. 

When  a  beam  deflects  under  the  action  of  a  load,  the  hori- 
zontal fibers  upon  one  side  of  the  neutral  surface  are  elongated 
and  upon  the  other  are  compressed.  The  internal  work  done 
will  be  found  by  taking  the  sum  of  the  products  formed  by 


172  APPENDIX   AND   TABLES.  CH.  VIII. 

multiplying  the  stress  upon  any  elementary  area  by  its  elonga- 
tion or  compression. 

Using  the  same  notation  as  in  Chapter  III.,  the  horizontal 
unit-stress  at  any  distance  z  from  the  neutral  axis  is  represented 

by  —  .     In  the  distance  dx  the  elongation  or  compression  due 

to  this  unit-stress,  is  by  (2)  found  to  be  —  —  .    The  elementary 

work  of  a  fiber-of  the  area  a  under  this  gradually  applied  unit- 
stress  hence  is, 

i    Saz        Szdx 


The  work  done  in  the  distance  dx  by  all  the  fibers  in  the  cross- 
section  now  is, 


S       M* 
Here  2asf  =  /and  from  formula  (4),  the  value  of  -7-  is  -— 

~,       ,  .,.          "'''' 

Therefore  dK  — 


. 

This  is  the  formula  for  the  work  done  in  the  distance  dx.  By 
expressing  M  as  a  function  of  x,  and  integrating,  the  total  in- 
ternal work  K  between  assigned  limits  can  be  found. 

For  example,  consider  a  cantilever  beam  loaded  at  the  end 
with  a  weight  P.  Here  M  =  —  Px.  Inserting  this  and  in- 
tegrating between  the  limits  o  and  /,  gives, 


for  the  total  int  rnal  work  in  the  beam  due  to  a  load  which  is 
gradually  applied. 


ART.  84.  THE   INTERNAL   WORK   IN   BEAMS.  1/3 

The  preceding  furnishes  a  new  method  of  deducing  the  de- 
flection of  a  beam  loaded  with  a  single  weight  P.  Let  A  be 
the  deflection  under  the  weight.  Then  \PA  is  the  external 
work  done  by  the  load  P  upon  the  beam,  and  this  must  equal 
the  internal  work  K.  Hence  the  formula, 

(18)  PA  = 


El  ' 

i/ 

from  which  A  may  be  found  for  particular  cases. 

For  example,  consider  a  cantilever  beam  loaded  at  the  end 

with  P.     Then  the  internal  work  is,  as  shown  above, . 

6EI 

Hence  the  deflection  A  is, 

which  is  the  same  as  otherwise  found  in  Art.  34. 

For  a  simple  be 

and  (18)  becomes, 

ri 

from  which  the  deflection  is, 


For  a  simple  beam  loaded  at  the  middle  the  value  of  Mis  — 


which  is  the  same  as  found  in  Art.  35  by  the  use  of  the  elastic 
curve. 

Prob.  139.  Prove  that  the  internal  work  caused  by  a  uni- 
formly distributed  load  on  a  cantilever  beam  is  ^-tns  of  that 
caused  by  the  same  load  applied  at  the  end. 

Prob.  140.  Deduce  by  the  method  of  Art.  35,  and  also  by 
the  use  of  the  principle  of  internal  work,  the  deflection  under 
a  load  P  which  is  placed  upon  a  simple  beam  at  a  distance  \l 
from  one  end. 


174  APPENDIX  AND  TABLES.  CH.  VIII. 

ART.  85.    ANSWERS  TO  PROBLEMS. 

Below  will  be  found  the  answers  to  about  nine-tenths  of  the 
problems  stated  in  the  preceding  pages,  the  number  of  the 
problem  being  in  parenthesis  and  the  answer  immediately  fol- 
lowing. It  has  been  thought  well  that  some  answers  should 
be  omitted  in  order  that  the  student  may  struggle  with  them 
to  ascertain  the  truth,  according  to  his  best  knowledge  of  the 
subject,  rather  than  to  make  his  numerical  results  agree  with 
given  figures.  However  satisfactory  it  may  be  to  the  student 
to  know  the  result  of  an  exercise  he  is  to  solve,  let  him  remem- 
ber that  after  commencement  day  the  answers  to  problems  will 
never  be  given. 

The  unit-stresses  to  be  employed  in  solutions  will  be,  unless 
otherwise  stated  in  the  problem,  uniformly  taken  from  the 
tables  given  in  the  text  and  in  Art.  86.  Considering  the  great 
variation  in  these  data  it  has  not  been  thought  best  to  carry 
the  numerical  answers  to  more  than  three  significant  figures, 
but  in  making  the  solution  four  significant  figures  should  be 
retained  through  the  work  in  order  that  the  third  may  be  correct 
in  the  final  result. 

Chapter  I.  (i)  7.2,  7.06,  and  86.4  square  inches.  (2)  173, 
34.7,  and  4320  pounds.  (3)  55000  pounds  per  square  inch. 
(4)  70000  pounds.  (5)  165000  pounds.  (6)  0.15  inches. 
(7)  25  ooo  ooo  pounds  per  square  inch.  (8)  0.004  inches. 
(9)  About  3^  inches  in  diameter,  (u)  2880  and  5400  feet. 
(12)  0.00153  inches.  (13)  52900  pounds  per  square  inch, 
(14)  849  pounds  per  square  inch.  (15)  About  if  inches  in  di- 
ameter. (16)  9  for  AB  and  23  for  BC. 

Chapter  II.  (17)  O.88  inches  if/=  15.  (18)  I  170  pounds 
per  square  inch.  (19)  2  500  pounds  per  square  inch.  (20)  i  620 
pounds  per  square  inch.  (22)  2^  inches.  (23)  57  per  cent ; 
/=  7.7.  (24)  3.28  inches;  about  0.73.  (25)  0.0032  inches. 


ART.  85.  ANSWERS   TO   PROBLEMS.  175 

Chapter  III.     (27)   z\  inches.     (29)   998   and   742    pounds. 

(30)  +  800,  +  i6o,and  —  180  at  i,  3,  and  5  feet  from  left  end. 

(31)  —  10,  —40,—  90,—  40,—  10  pound-feet.    (33)  Y=  140  and 
X=2&  pounds.      (34)  2700  pounds.      (35)  X  —  -  Z=  375 
pounds.     (36)  About   27.     (37)    4.20  inches.      (38)  c  =  1.714 
inches,  /=  7.39  inches4.     (39)  \bd*  and  -fabd*.     (41)  At  5.37 
feet  from  left  end  ;  M  —  689  pound-feet.     (42)  No.     (43)  The 
bar  will  break.     (44)  294  pounds  per  linear  foot.     (45)  About 
6000  pounds.    (47)  8.87  inches.   (48)  0.0178  inches.   (49)  About 
615  pounds.     (51)  Ultimate  strengths   about    as  4  to  I,  while 
working  strengths  for  a  steady  load  are  about.  as   1.8  to   i. 
(52)  3.7  and  1.8.     (53)  7  feet,  8  inches.     (54)  The  beam  will 
break.     (56)  J>  =  5560    and   S'  =  3  410    pounds    per    square 
inch.     (57)  209  inches,  418  inches,  and  oo.     (58)  0.622  inches. 
(59)   14  500  ooo  pounds  per  square  inch.  (62)  As  8  to  3  ;  as  64 

to  9.  (63)  7^  inches.  (64)  0.243  inches.  (65)  x  =  6  ooo  —  -p 
for  the  first  case  ;  the  shear  at  supports  is  independent  of  x. 
(68)  0.72  inches. 

Chapter  IV.  (69)  The  diagrams  should  always  be  drawn  on 
cross-section  paper.  (70)  4  and  3^.  (72)  k  =  0.366,  and 
k  —  0.577.  (73)  /  =  2.828  m.  (74)  5  =  829  pounds  per  square 
inch.  (76)  A  heavy  1  5-inch  beam;  a  light  1  5-inch  beam. 
(78)  0.0269  inches.  (80)  R,  =  R4  =  T\w/;  X,=R9  =  \\wL 


(81)  —  o°  25'  47".    (83)  -  =  7.2  which  requires  the  light  6-inch 
beam.     (84)  —-fowl.     (86)  n  —0.6095. 

Chapter  V.  (88)  9.15  inches.  (89)  2  inches.  (90)  205000 
pounds.  (92)  69.7  tons.  (93)  5.05  inches.  (94)  ^  =  40900, 
£=irsW  (96)  r  =  0.84  inches.  (97)  if.  (98)  2.35  and  24. 
(99)  250000  pounds.  (101)  13^  and  i6£  inches  square. 
(104)  Draw  Fig.  48  so  as  to  make  bq  =  o  ;  then  state  equation 
of  moments  and  reduce  it  by  the  relation  between  the  similar 
triangles. 


i;6  APPENDIX  AND   TABLES.  CH.  VIII. 

Chapter  VI.  (106)  30  pounds.  (107)  105  degrees.  (108) 
720,  270,  and  290  pound-inches.  (109)  I  876  pounds  per  square 
inch,  (i  10)  7  =  0.0361^*  and  £  =  0.577^.  (111)7=26.5 
and  £=  3.41.  (112)  i  680  pounds.  (113)  9  380000  pounds  per 
square  inch.  (114)  64  horse-power.  (115)9.7.  (116)2.65  and 
3.58  inches.  (117)  32  and  8.  (119)  As  v^  to  3.  (120)  5  140 
and  8  560  pounds  per  square  inch. 

Chapter  VII.  (122)  3  720  pounds  per  square  inch.  (123)4690 
pounds  per  square  inch.  (124)  The  light  9-inch  beam. 
(125)  Nearly  8  inches.  (126)  9  inches.  (127)  £  =  9420, 
0=54°  20';  5=7160,  0  =  9°  20'.  (129)  5.4.  (130)  2$ 
inches.  (131)  S=  5  660  and  Ss  —  202  pounds  per  square  inch. 
(132)  At  3  inches  from  neutral  surface  5  =  2000,  Sh  =  250,  and 
/  =  2  030  pounds  per  square  inch. 

ART.  86.    TABLES  OF  CONSTANTS. 

The  following  tables  recapitulate  the  mean  values  of  the  con- 
stants of  the  strength  of  materials  which  have  been  given  in 
the  preceding  pages.  It  is  here  again  repeated  that  these 
values  are  subject  to  wide  variations  dependent  on  the  kind 
and  quality  of  the  material,  and  for  many  other  reasons.  Tim- 
ber, for  instance,  varies  in  strength  according  to  the  climate 
where  grown,  the  soil,  the  age  of  the  tree,  the  season  of  the 
year  when  cut,  the  method  and  duration  of  the  process  of 
seasoning,  the  part  of  the  tree  used,  the  knots  and  wind  shakes, 
the  form  and  size  of  the  test  specimen,  and  the  direction  of  its 
fibers,  so  that  it  is  a  difficult  matter  to  state  definite  numerical 
values  concerning  its  elasticity  and  strength.  The  quality  of 
the  material  causes  a  yet  wider  variation,  so  wide  in  fact  that 
in  some  cases  testing  machines  alone  could  scarcely  distinguish 
between  wrought  iron  and  steel ;  for  while  the  higher  grades  of 
steel  have  much  greater  strength  than  the  tables  give,  the  mild 
structural  and  merchant  steels  may  have  values  almost  as  low 


ART.  86. 


TABLES   OF  CONSTANTS. 


as  the  average  constants  for  wrought  iron.  In  general,  there- 
fore, the  following  values  should  not  be  used  m  act uaT  cases  of 
investigation  and  design  except  for  approximate  computations. 

Detailed  tables  giving  the  results  of  experiments  upon 
numerous  kinds  and  qualities  of  materials  may  be  found  in  the 
following  books. 

WOOD'S  Resistance  of  Materials ;  New  York,  1880. 

THURSTON'S  Materials  of  Engineering;  New  York,  1884. 

TRAUTWINE'S  Engineers'  Pocket  Book;  New  York,  1885. 

LANZA'S  Applied  Mechanics;  New  York,  1885. 

UNWIN'S  Testing  of  Materials;  London,  1888. 

BURR'S  Elasticity  and  Strength  of  Materials ;  New  York,  1888. 

TABLE  I. 


Mean  Weight. 

Coefficient  of  Linear  Expansion. 

Material. 

Pounds  per 
cubic  foot. 

Kilograms  per 
cubic  meter. 

For  i°  Fah. 

For  iff  Cent. 

Timber, 

40 

600 

O.OOOOO2O 

0.0000036 

Brick, 

"5 

2  000 

0.0000050 

0.0000090 

Stone, 

1  60 

2  560 

0.0000050 

0.0000090 

Cast  Iron, 

450 

7  200 

O.OOOOO62 

O.OOOOII2 

Wrought  Iron, 

480 

7  700 

0.0000067 

O.OOOOI2I 

Steel, 

490 

7  800 

0.0000065 

0.0000117 

TABLE  II. 


Elastic  Limit. 

Coefficient  of  Elasticity. 

Material. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms  per 
square  centimeter. 

Timber, 

3  ooo 

210 

I  500  ooo 

105  ooo 

Cast  Iron, 

6  ooo 

420 

15  ooo  ooo 

i  050  ooo 

Wrought  Iron, 

25  ooo 

I  750 

25  ooo  ooo 

i  750  ooo 

Steel, 

50000 

3  500 

30  ooo  ooo 

2  IOO  OOO 

APPENDIX  AND  TABLES. 

TABLE  III. 


CH.  VIII. 


Material. 

Ultimate  Tensile  Strength. 

Ultimate  Compressive  Strength. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Timber, 

10  000 

700 

8  000 

560 

Brick, 

2OO 

14 

2  5OO 

175 

Stone, 

6  OOO 

420 

Cast  Iron, 

2O  OOO 

I  400 

90  ooo 

6  300 

Wrought  Iron, 

55000 

3850 

55000 

3  850 

Steel,  . 

100  000 

7  ooo 

150000 

10  500 

TABLE  IV. 


Ultimate  Shearing  Strength. 

Modulus  of  Rupture. 

Material. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Pounds  per 
square  inch. 

Kilograms 
per  square 
centimeter. 

Timber, 

|         600    ) 
(      3000    f 

\            42    I 
\            210     J 

9  OOO 

630 

Stone, 

2  OOO 

140 

Cast  Iron, 

20000 

I  400 

35000 

2450 

Wrought  Iron, 

50000 

3  500 

55000 

3850 

Steel, 

70  ooo 

4900 

1  2O  OOO 

8  400 

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